SFNQuestions Posted February 22, 2017 Share Posted February 22, 2017 Like if I have a parabola of velocity versus time or position versus time, is there any physical meaning to the arc length of that function? Link to comment Share on other sites More sharing options...
Bender Posted February 22, 2017 Share Posted February 22, 2017 Changing the scale, or the units, will change the arc length, so no. It would also require adding different units. Link to comment Share on other sites More sharing options...
fiveworlds Posted February 22, 2017 Share Posted February 22, 2017 Like if I have a parabola of velocity versus time or position versus time, is there any physical meaning to the arc length of that function? You take a bus from your home to school/work along a curved road. If you leave your house at 8:20 and arrive at 9:10. What is the arc length?? Link to comment Share on other sites More sharing options...
imatfaal Posted February 23, 2017 Share Posted February 23, 2017 Changing the scale, or the units, will change the arc length, so no. It would also require adding different units. But your argument would apply to the area under the curve as well - and there is most definitely a very useful physical interpretation to that. I cannot see a useful meaning to the arc-length yet - but yours is not a valid counterargument Like if I have a parabola of velocity versus time or position versus time, is there any physical meaning to the arc length of that function? The dimension of the area under a velocity time graph is length - although I think I have phrased that awfully it makes sense. What is the dimension of the arc-length of a velocity time graph? 1 Link to comment Share on other sites More sharing options...
Bender Posted February 23, 2017 Share Posted February 23, 2017 But your argument would apply to the area under the curve as well - and there is most definitely a very useful physical interpretation to that. I cannot see a useful meaning to the arc-length yet - but yours is not a valid counterargument I think my counterargument works just fine. The surface area is a multiplication and, in case of velocity and time, you get a distance with matching units. Changing the units will change the unit and value of the distance (whatever the units of time and velocity you use, the resulting unit will always be one of distance), but not the distance itself. Link to comment Share on other sites More sharing options...
steveupson Posted February 23, 2017 Share Posted February 23, 2017 The dimension of the area under a velocity time graph is length - although I think I have phrased that awfully it makes sense. What is the dimension of the arc-length of a velocity time graph? Right. Velocity x time produces a different result than speed x time. I raised this question (I think it's the same question) in the dimensional analysis thread. I was told there that even though the two things are different mathematically, they have the same dimension (I think that's what I was told.) The mainstream view on this question is defined that way, but it isn't explained why, at least not anywhere that I've been able to find. It doesn't make sense mathematically or conceptually. There is an alternative view that does make sense, both mathematically and conceptually. The way it should work is that since velocity is displacement over time, instead of length (arc-length in this case) over time, then there should be a velocity time graph where displacement instead of length is the area under the curve. This difference isn't as subtle as it sounds. In order to quantify displacement we have to know direction because displacement is length x direction. I can't find any evidence that anyone has seriously proposed considering direction as a base quantity on a par with length or time. This looks like a glaring omission. We know that the linear expression of spherical excess (E = A + B + C - π) isolates a quantity, and yet some insist that this quantity that is being isolated isn't a real quantity. Hopefully I won't receive bullying from the staff for publicly talking to you about this. I've been warned many times that this is a taboo subject, but you did ask. I think the question being asked publicly by another member should be grounds for me to offer a comment, even though I've been previously banned from articulating my opinion on this subject. Link to comment Share on other sites More sharing options...
Bender Posted February 23, 2017 Share Posted February 23, 2017 (edited) Right. Velocity x time produces a different result than speed x time. I raised this question (I think it's the same question) in the dimensional analysis thread. I was told there that even though the two things are different mathematically, they have the same dimension (I think that's what I was told.) The mainstream view on this question is defined that way, but it isn't explained why, at least not anywhere that I've been able to find. It doesn't make sense mathematically or conceptually. There is an alternative view that does make sense, both mathematically and conceptually. The way it should work is that since velocity is displacement over time, instead of length (arc-length in this case) over time, then there should be a velocity time graph where displacement instead of length is the area under the curve. This difference isn't as subtle as it sounds. In order to quantify displacement we have to know direction because displacement is length x direction. I can't find any evidence that anyone has seriously proposed considering direction as a base quantity on a par with length or time. This looks like a glaring omission. We know that the linear expression of spherical excess (E = A + B + C - π) isolates a quantity, and yet some insist that this quantity that is being isolated isn't a real quantity. Hopefully I won't receive bullying from the staff for publicly talking to you about this. I've been warned many times that this is a taboo subject, but you did ask. I think the question being asked publicly by another member should be grounds for me to offer a comment, even though I've been previously banned from articulating my opinion on this subject. I'm not entirely sure what you are talking about, but we already have a way of addressing direction: vectors. The area under the curve does not contain a direction for two reasons: - it is a scalar, not a vector and there would be no difference regardless what the direction of the velocity is - it is not the displacement (which is the difference between end point and point of origin, regardless of the path taken); but rather the distance travelled (which does take into account the path taken). The distance travelled is not a vector, because the path can go along different directions. Edited February 23, 2017 by Bender Link to comment Share on other sites More sharing options...
steveupson Posted February 23, 2017 Share Posted February 23, 2017 I'm not entirely sure what you are talking about, but we already have a way of addressing direction: vectors. Yes, true, but vectors are ratios, not numbers. And there's a special treatment in this particular case of speed. It isn't like any of the other vectors. The x,y,z components of a vector contain a ratio that describes a direction in Euclidean 3-space. This ratio is pretty much meaningless in spacetime. The reason it's meaningless is that the speed of light is constant, and therefore length isn't a constant (it's simply a base quantity) and it varies in a manner that is inversely proportional to time. "Vectors have magnitude and direction, scalars only have magnitude. The fact that magnitude occurs for both scalars and vectors can lead to some confusion. There are some quantities, like speed, which have very special definitions for scientists. By definition, speed is the scalar magnitude of a velocity vector." https://www.grc.nasa.gov/www/k-12/airplane/vectors.html We also have another method of addressing direction: angles. The same is true for this treatment. An angle is a ratio that exists between the diameter and circumference of a circle. Once again direction is expressed as a ratio and not as a quantity. Spherical excess expresses direction as a quantity, a number, not as a ratio. The relationship between a flat plane (Euclidean 2-space) and the surface of a sphere can be expressed as a real quantity. By this I mean that it can be expressed as a scalar value. This is a real quantity, or at least I believe that it meets the technical definition of one since it's a number and not a ratio. If spherical excess isn't a quantity then what is it? There's another new way to express the quantity that manifests spherical excess. This new method permits the curvature of spacetime to be linearized. If we call the traditional method of quantifying spherical excess the linear method then we could refer to the new method as the circular method. It produces more information than the traditional method. Basically, what it shows is that space that isn't curved (no spherical excess) is dominated by the sine function while space with maximum curvature (maximum spherical excess) is dominated by the hyperbolic function. By the term linearized I mean that since a smooth function exists between a sine and a hyperbola, there is a way to scale the curvature. The area under the curve does not contain a direction for two reasons: - it is a scalar, not a vector and there would be no difference regardless what the direction of the velocity is Again, explain how this gets resolved in the twin quasar question. Thegravitational lensing associated with the twin quasar shows this to be incorrect to a certainty. "30 years of observation made it clear that image A of the quasar reaches earth about 14 months earlier than the corresponding image B, resulting in a difference of path length of 1.1 ly." https://en.wikipedia...iki/Twin_Quasar By definition, the light from A has a higher velocity than the light from B, even though they have the same speed. There should be three separate values with different graphs; velocity of light for image A, velocity of light for image B, and a third velocity c for speed of light (based on the geodesic between the quasar and the earth.) - it is not the displacement (which is the difference between end point and point of origin, regardless of the path taken); but rather the distance travelled (which does take into account the path taken). The distance travelled is not a vector, because the path can go along different directions. Yes. It should account somehow for the path taken, but it really doesn't. I think we can agree that the path can go in different directions, we disagree on what that means. My understanding is that spacetime is a combination of two different things, time and space. I also understand that space is also a combination of two different things, direction and distance. Spacetime has three base quantities, time, length, and direction. This is different than saying that spactime has two base quantities, time and length and the ratio between lengths. Link to comment Share on other sites More sharing options...
studiot Posted February 24, 2017 Share Posted February 24, 2017 To answer the original question, if the OP is still interested; Lines can exist in two or three dimensions so To find real world examples of use of the length of arc look for things that are linelike. In 2D say you are constructing a suspension bridge or a posttensioned bridge. Bridge cables are expensive, per metre, so the bridge owner only wants to pay for the cable actually used. The bridge builder on the other hand need to know exactly how much cable to order, so that he can do the work. So both have a strong interest in the exact length of cable. In 3D the manufacturer of electric windings needs to know the length of wire to complete the number of turns on a bobbin. As does a coil spring manufacturer need the exact length of spring wire for his coils. Those were all physical examples. Navigators and surveyors need to know the arc length on the curved surface of the Earth, a theoretical, but nevertheless important value. For a financial example related to plots, say you were saving money, at regular intervals. A the length of a plot of money v time would tell you how much you had at any given time. Link to comment Share on other sites More sharing options...
Bender Posted February 24, 2017 Share Posted February 24, 2017 "Vectors have magnitude and direction, scalars only have magnitude. The fact that magnitude occurs for both scalars and vectors can lead to some confusion. There are some quantities, like speed, which have very special definitions for scientists. By definition, speed is the scalar magnitude of a velocity vector." https://www.grc.nasa.gov/www/k-12/airplane/vectors.html Again, explain how this gets resolved in the twin quasar question. Thegravitational lensing associated with the twin quasar shows this to be incorrect to a certainty. "30 years of observation made it clear that image A of the quasar reaches earth about 14 months earlier than the corresponding image B, resulting in a difference of path length of 1.1 ly." https://en.wikipedia...iki/Twin_Quasar By definition, the light from A has a higher velocity than the light from B, even though they have the same speed. There should be three separate values with different graphs; velocity of light for image A, velocity of light for image B, and a third velocity c for speed of light (based on the geodesic between the quasar and the earth.) (emphasis added to what I'm replying to) What definition are you talking about? You seem to agree that speed is the magnitude of velocity, then you say that one beam has a higher velocity, while the speed is the same. How can two vectors have different magnitudes if their magnitude is the same? Link to comment Share on other sites More sharing options...
fiveworlds Posted February 24, 2017 Share Posted February 24, 2017 Lets say we know how fast a person's heart beats. The machine will tell you if it beats to fast or too slow but what does the arc-length tell you?? Link to comment Share on other sites More sharing options...
Bender Posted February 24, 2017 Share Posted February 24, 2017 On itself: nothing. If you normalise it by dividing by e.g. total time and average heart rate, you get a value that says something about the variability. I guess this works for any graph, but I think there are better ways to describe variability. Link to comment Share on other sites More sharing options...
fiveworlds Posted February 24, 2017 Share Posted February 24, 2017 If you normalise it by dividing by e.g. total time and average heart rate, you get a value that says something about the variability So arc-length is some form of a measure of how variable the graph is. In the case of heart rate it could indicate arrhythmia. On itself: nothing. When do you ever have the arc-length by itself?? Link to comment Share on other sites More sharing options...
swansont Posted February 24, 2017 Share Posted February 24, 2017 For a financial example related to plots, say you were saving money, at regular intervals. A the length of a plot of money v time would tell you how much you had at any given time. Would it? Take a plot, and plot, and invert it. Same arc length. Vastly different results for the end value. The only information I have been able to come up with is this: the minimum length is for a constant function. If the arc length is not the minimum, the function is not a constant. edit: there also may be other min/max or optimal/not optimal situations that could be present. Link to comment Share on other sites More sharing options...
steveupson Posted February 24, 2017 Share Posted February 24, 2017 (emphasis added to what I'm replying to) What definition are you talking about? You seem to agree that speed is the magnitude of velocity, then you say that one beam has a higher velocity, while the speed is the same. How can two vectors have different magnitudes if their magnitude is the same? In spacetime that is not curved (as with Euclidean 3-space) the two definitions are identical. It's only when we consider curved spacetime that the issue arises. The problem with speed being the magnitude of velocity is that in curved spacetime the velocity of light can be any value less than the speed of light. The two are not the same in nature (curved spacetime) and we accept speed as the constant, not velocity, by convention. In the twin quasar example we have one change of position (or displacement) and three different times. The magnitude of the velocity for the straight-line or geodesic path is [latex]c[/latex] which is the rate of change of position of light. The magnitude of the velocity for image B is different because the rate is [latex]c- n_1[/latex], where [latex]n_1[/latex] is the difference in velocities caused by the time difference due to path difference. The magnitude for image A is [latex]c - (n_1 + n_2)[/latex] where [latex]n_2[/latex] is the change in velocity due to the 1.1 year delay caused by the distance traveled for image B being 1.1 ly further than that traveled by image A. Or, in order for the velocities to be identical for all three times, the displacement between the quasar and earth must be different for each path. That doesn't really fit the definitions that we're using. On a similar note, there are other derived units that are ambiguous in the same way because they use [latex]l^3[/latex] as a volume. Once again, this loses any meaning at all in curved spacetime for similar reasons. In curved space [latex]l^3[/latex] becomes ambiguous because length, once again, is not a constant. The only information I have been able to come up with is this: the minimum length is for a constant function. If the arc length is not the minimum, the function is not a constant. Yes. The velocity of light in curved spacetime is not a constant. The speed of light is a constant and is represented by the shortest length. I'm using the term geodesic and I hope that is a proper use of the term. Link to comment Share on other sites More sharing options...
Bender Posted February 24, 2017 Share Posted February 24, 2017 So arc-length is some form of a measure of how variable the graph is. In the case of heart rate it could indicate arrhythmia.It is possible that such a correlation exists, if you take a normalised arclength. There are several ways to do that. When do you ever have the arc-length by itself??When you don't normalise it. If you e.g. compare two arclengths of two graphs with different time intervals, you'll get nothing useful. Link to comment Share on other sites More sharing options...
swansont Posted February 24, 2017 Share Posted February 24, 2017 Yes. The velocity of light in curved spacetime is not a constant. The speed of light is a constant and is represented by the shortest length. I'm using the term geodesic and I hope that is a proper use of the term. We're talking about a graph. Link to comment Share on other sites More sharing options...
steveupson Posted February 24, 2017 Share Posted February 24, 2017 (edited) We're talking about a graph. From the OP, specifically we're talking about a graph of velocity vs time or position vs time. I think it's appropriate to assume spacetime as part of the discussion. on edit>>> An example would be that the plot of position or velocity vs time of a race car around a circular track will be a sine wave. The plots of a race car down a drag strip will be a parabola. When speed is the constant (as with the speed of light) then the plots of the drag strip are a line and they are the plots that represent this constant. Edited February 24, 2017 by steveupson Link to comment Share on other sites More sharing options...
swansont Posted February 24, 2017 Share Posted February 24, 2017 From the OP, specifically we're talking about a graph of velocity vs time or position vs time. I think it's appropriate to assume spacetime as part of the discussion. No, spacetime doesn't matter here. At all. You are overthinking the problem, and/or forcing some rather unusual ideas onto the problem. Link to comment Share on other sites More sharing options...
zztop Posted February 24, 2017 Share Posted February 24, 2017 Right. Velocity x time produces a different result than speed x time. I raised this question (I think it's the same question) in the dimensional analysis thread. I was told there that even though the two things are different mathematically, they have the same dimension (I think that's what I was told.) The mainstream view on this question is defined that way, but it isn't explained why, at least not anywhere that I've been able to find. It doesn't make sense mathematically or conceptually. There is an alternative view that does make sense, both mathematically and conceptually. The way it should work is that since velocity is displacement over time, instead of length (arc-length in this case) over time, then there should be a velocity time graph where displacement instead of length is the area under the curve. This difference isn't as subtle as it sounds. In order to quantify displacement we have to know direction because displacement is length x direction. I can't find any evidence that anyone has seriously proposed considering direction as a base quantity on a par with length or time. This looks like a glaring omission. We know that the linear expression of spherical excess (E = A + B + C - π) isolates a quantity, and yet some insist that this quantity that is being isolated isn't a real quantity. Hopefully I won't receive bullying from the staff for publicly talking to you about this. I've been warned many times that this is a taboo subject, but you did ask. I think the question being asked publicly by another member should be grounds for me to offer a comment, even though I've been previously banned from articulating my opinion on this subject. There are a lot of errors in your post: 1. There is no such thing as "velocity x time". What exists is [latex]\frac{d \vec{r}}{dt}=\vec{v}[/latex] 2. No knowledgeable person claims that "velocity is arc-length over time", you need to stop creating strawmen. 3. "the velocity of light can be any value less than the speed of light" is absolute nonsense since the "value" of the light velocity is ....its speed. And the speed of light is....c. You need to LEARN before you post. So far, you have not learned anything. Link to comment Share on other sites More sharing options...
steveupson Posted February 24, 2017 Share Posted February 24, 2017 There are a lot of errors in your post: 1. There is no such thing as "velocity x time". What exists is [latex]\frac{d \vec{r}}{dt}=\vec{v}[/latex] 2. No knowledgeable person claims that "velocity is arc-length over time", you need to stop creating strawmen. 3. "the velocity of light can be any value less than the speed of light" is absolute nonsense since the "value" of the light velocity is ....its speed. And the speed of light is....c. You need to LEARN before you post. So far, you have not learned anything. You're probably right and I'm probably wrong. In any case, can you plot the velocity of image A vs time and image B vs time? Or even better, can you plot the displacement of image A vs time and image B vs time? Assume that the twin quasar is 7,800,000,000 ly distant from earth. Since the displacement is the same, and the magnitude of the velocity is the same, by your methods the the two graphs should be identical, shouldn't they? Teach me. I don't understand your method. I do want to learn. No, spacetime doesn't matter here. At all. You are overthinking the problem, and/or forcing some rather unusual ideas onto the problem. How I think about this doesn't really have any effect on the physics. These ad hominem arguments from you are becoming wearisome. Link to comment Share on other sites More sharing options...
zztop Posted February 24, 2017 Share Posted February 24, 2017 (edited) You're probably right and I'm probably wrong. In any case, can you plot the velocity of image A vs time and image B vs time? Or even better, can you plot the displacement of image A vs time and image B vs time? Assume that the twin quasar is 7,800,000,000 ly distant from earth. Since the displacement is the same, and the magnitude of the velocity is the same, by your methods the the two graphs should be identical, shouldn't they? Teach me. I don't understand your method. I do want to learn. No, you do not want to learn, what you want to do is to continue posting nonsense. If you wanted to learn, you would have started by taking a break from posting and by taking a class in vector calculus. In the twin quasar example we have one change of position (or displacement) and three different times. The magnitude of the velocity for the straight-line or geodesic path is which is the rate of change of position of light. The magnitude of the velocity for image B is different because the rate is , where is the difference in velocities caused by the time difference due to path difference. There is no such thing as [latex]c-n_1[/latex]. Light speed doesn't add/subtract. The magnitude for image A is where is the change in velocity due to the 1.1 year delay caused by the distance traveled for image B being 1.1 ly further than that traveled by image A. There is no such thing as [latex]c-(n_1+n_2)[/latex].Light speed doesn't add/subtract. Edited February 24, 2017 by zztop Link to comment Share on other sites More sharing options...
swansont Posted February 24, 2017 Share Posted February 24, 2017 How I think about this doesn't really have any effect on the physics. These ad hominem arguments from you are becoming wearisome. There was no ad hominem argument. Link to comment Share on other sites More sharing options...
steveupson Posted February 24, 2017 Share Posted February 24, 2017 There was no ad hominem argument. I must have misunderstood. I thought your post was about me. Is there some other content that I missed? Link to comment Share on other sites More sharing options...
swansont Posted February 24, 2017 Share Posted February 24, 2017 I must have misunderstood. I thought your post was about me. Is there some other content that I missed? No, what you are missing is a working knowledge of the definition of an ad hominem argument. Link to comment Share on other sites More sharing options...
Recommended Posts
Create an account or sign in to comment
You need to be a member in order to leave a comment
Create an account
Sign up for a new account in our community. It's easy!
Register a new accountSign in
Already have an account? Sign in here.
Sign In Now