asprung Posted March 14, 2008 Posted March 14, 2008 As time slows down as a body approaches the speed of light and its length shortens does its other dimensions also change? How would this effect a space traveler whose mass should not change?
Reaper Posted March 14, 2008 Posted March 14, 2008 Only it's length changes. A space traveler's mass would also change, due to E = (1/(1-v2/c2)1/2 - 1) mc2 and m = m0/(1-v2/c2)1/2 (e.g. It gains more mass due to it's relativistic speed and large kinetic energy).
NeonBlack Posted March 14, 2008 Posted March 14, 2008 "Reaper"- You are correct that only the length in the direction of motion changes, but mass does not change as a function of speed.
swansont Posted March 14, 2008 Posted March 14, 2008 Only it's length changes. A space traveler's mass would also change, due to E = (1/(1-v2/c2)1/2 - 1) mc2 and m = m0/(1-v2/c2)1/2 (e.g. It gains more mass due to it's relativistic speed and large kinetic energy). The "m" in your first equation should be the rest mass. And the use of relativistic mass usually just adds confusion. The rest mass is unchanged. ——— In the space traveler's own frame, none of his attributes change, since the astronaut observes himself to be at rest.
asprung Posted March 14, 2008 Author Posted March 14, 2008 I am confused. If the space traveler's length in the direction of travel would diminish and his mass would not decrease then his body density would, which should kill him. Yet in his own frame none of this really happens. Does it or doesn’t it.
thedarkshade Posted March 14, 2008 Posted March 14, 2008 Three things happen when traveling: 1.Experience less time 2.Contract 3.Become heavier For the two later ones you can calculate yourself by: [math]l=l_0\sqrt{1-\beta^2}[/math] .... Lorentz Contraction where [math]l_0[/math] is rest length and [math]m=\frac{m_0}{\sqrt{1-\beta^2}}[/math] where [math]m_0[/math] is rest mass!
swansont Posted March 14, 2008 Posted March 14, 2008 Three things happen when traveling:1.Experience less time 2.Contract 3.Become heavier For the two later ones you can calculate yourself by: [math]l=l_0\sqrt{1-\beta^2}[/math] .... Lorentz Contraction where [math]l_0[/math] is rest length and [math]m=\frac{m_0}{\sqrt{1-\beta^2}}[/math] where [math]m_0[/math] is rest mass! These things are what are seen by the stationary observer, not the traveler. And the third item is, by convention, not often used. Rest mass is used, as is the total energy. I am confused. If the space traveler's length in the direction of travel would diminish and his mass would not decrease then his body density would, which should kill him. Yet in his own frame none of this really happens. Does it or doesn’t it. "Does it or doesn't it" isn't a valid question for these things. What is observed depends on your frame of reference. Asking if the length actually contracts assumes there is an absolute reference frame that would allow you to answer that, and such a frame does not exist. The contraction depends on your reference frame.
NeonBlack Posted March 15, 2008 Posted March 15, 2008 And what's so hard about calculating time dilation? If you're not willing to let go of the idea of "relativistic mass," consider this: There are actually 2 relativistic masses, which depend on direction (yes, you read that right!) The "longitudinal mass," in the direction of motion is [math]m_l=\frac{m_0}{1-\beta^2}[/math] And the "transverse mass," perpendicular to the direction of motion is [math]m_t=\frac{m_0}{\sqrt{1-\beta^2}}[/math] This is crazy. However, none of this becomes neccessary if you just accept that mass is an invariant. Not only is this confusing, but it is just wrong. If you derive the equations in SR, you will see that all the factors of gamma come from time dilation and have nothing to do with changing mass.
thedarkshade Posted March 15, 2008 Posted March 15, 2008 If you derive the equations in SR, you will see that all the factors of gamma come from time dilation and have nothing to do with changing mass. Suppose we travel at 0.8c, and have a mas of 50kg. So: [math]f=\sqrt{1 - \beta^2}=\sqrt{1-(0.8)^2}=\sqrt{0.36}=0.6[/math] .. do [math]\sqrt{1-\beta^2[/math]=0.6 and gamma function: [math]\gamma = \frac{1}{0.6}=1.666666......[/math] then mass: [math]m=\frac{m_0}{\sqrt{1-\beta^2}}=\frac{50}{0.6}=83.333333...[/math] and there is some relation: [math]\frac{m}{m_0}=\gamma[/math] [math]\frac{83.33333.....}{50}=1.666666...[/math] [math]1.6666....=1.6666....[/math] See!
swansont Posted March 15, 2008 Posted March 15, 2008 You're missing the point. You can make the math work, because the same factor (gamma) is appearing in the equations. However, that doesn't make it correct. Let's take the example of someone standing on a scale on some planet, and there is a pointer on a dial that indicates the weight. We then transform into a reference frame where the planet is moving at some relative speed. Does the indicator move to a new number as we increase v? The answer, of course, is "No, it can't!" because we have to observe the same event — the dial can't point to 500 N in one frame but 1000 N in another. Even thought he contention is that both the mass of the person and the planet should have increased, making the gravitational attraction increase (as the square of gamma, since you have multiplied the mass terms) You won't observe the distance of a binary star orbit (not in the direction of length contraction) change, either, when the speed increases, for the same reason. Relativistic mass is a sloppy shortcut that breaks relativity if it's actually applied broadly. That it gives the right answer in a subset of problems doesn't make it right.
asprung Posted March 16, 2008 Author Posted March 16, 2008 Is the space traveler's body physically compressed in length or is this an illusion from the observers frame. I do not see how the space traveler can be standing at his full height and at the same time be compressed. He only has one body.
iNow Posted March 16, 2008 Posted March 16, 2008 Is the space traveler's body physically compressed in length or is this an illusion from the observers frame. I do not see how the space traveler can be standing at his full height and at the same time be compressed. He only has one body. It's not an illusion. He really is length contracted, but only relative to an observer. Relative to himself, everything about him is length contracted, including the molecules and atoms which compose him and his body. So, since the ratios are the same (the space between all of those atoms is ALSO length contracted), then they notice no difference. Compression is not really the correct term, as it implies making something denser and decreasing the amount/ratio of space between each molecule and atom. However, since the molecules and atoms are also being length contracted, the ratio of space between them remains the same. This is why it's called length contraction, not compression. I used this example in another thread recently. If you have a stick which expands from 1 meter to 10 meters, but the ruler you are using to measure that stick also expands by 10x at the same time, then the measurement is not effected because the ratio is unchanged.
swansont Posted March 16, 2008 Posted March 16, 2008 It's not an illusion. He really is length contracted, but only relative to an observer. Relative to himself, everything about him is length contracted, including the molecules and atoms which compose him and his body. So, since the ratios are the same (the space between all of those atoms is ALSO length contracted), then they notice no difference. No, nothing about the observer is contracted in his own frame. It's not a matter of "noticing" a difference. Contraction only happens when observing something in a different reference frame. I agree with the first sentence — it's not an illusion. What you measure is what you measure. It is because measurements like this that are relative, that it's useful to find invariant quantities, like rest mass or 4-vector quantities like spactime interval (which is invariant because time dilation and length contraction effects balance)
iNow Posted March 16, 2008 Posted March 16, 2008 No, nothing about the observer is contracted in his own frame. It's not a matter of "noticing" a difference. Contraction only happens when observing something in a different reference frame. I agree with the first sentence — it's not an illusion. What you measure is what you measure. It is because measurements like this that are relative, that it's useful to find invariant quantities, like rest mass or 4-vector quantities like spactime interval (which is invariant because time dilation and length contraction effects balance) Well, okay. That is exactly what I intended to suggest, but looking back I can see that this point could easily have been lost in the shorthand I chose. Stupid red wine. Thanks for clarifying, for both me and others.
asprung Posted March 16, 2008 Author Posted March 16, 2008 What you are saying is that the space between the molecules is really diminished but only for the observer not for him. How can that be? He only has one body. If this space is diminished but only in length his body would become distorted doing who knows what damage to him.
iNow Posted March 16, 2008 Posted March 16, 2008 Correction: Relative to the observer, spacetime itself is contracted for the traveller.
Klaynos Posted March 16, 2008 Posted March 16, 2008 What you are saying is that the space between the molecules is really diminished but only for the observer not for him. How can that be? He only has one body. If this space is diminished but only in length his body would become distorted doing who knows what damage to him. Not just the space between the molecules, but the molecules themselves appear smaller, as do the size of atoms... it's all relative :|
asprung Posted March 16, 2008 Author Posted March 16, 2008 Do the molecules, the space between them, Etc. really become smaller or merely appear to. It’s hard to conceive of molecules etc. becoming smaller in one direction which would change as the space traveler moves around. It should be rough on his body.
swansont Posted March 16, 2008 Posted March 16, 2008 Do the molecules, the space between them, Etc. really become smaller or merely appear to. It’s hard to conceive of molecules etc. becoming smaller in one direction which would change as the space traveler moves around. It should be rough on his body. Again, you can't ask what "really" happens, because this assumes there is an absolute reference frame. Events happen, e.g. whether or not there was a collision has to be true in all frames. But a measurement of dimension or time is relative, as is any quantity that is not invariant. Time and length are not invariant quantities.
asprung Posted March 25, 2008 Author Posted March 25, 2008 If the space travelers is length contracted his body would become distorted. I dont know why I cant ask if this really happens. He only has one body,and something really happens to it or doesnt.
Cap'n Refsmmat Posted March 25, 2008 Posted March 25, 2008 That's what'd make sense, but that's not the case. Length depends on who's measuring it. It seems incredibly strange (and contrary to common sense) but it's what happens.
swansont Posted March 25, 2008 Posted March 25, 2008 That's what'd make sense, but that's not the case. Length depends on who's measuring it. It seems incredibly strange (and contrary to common sense) but it's what happens. And you have to take this one step further — not only do length and time depend on who is measuring it, but there is no way to determine that any frame of reference gives he "right" answer; nobody can physically determine if they are moving or at rest, if the motion is inertial.
thedarkshade Posted March 25, 2008 Posted March 25, 2008 And you have to take this one step further — not only do length and time depend on who is measuring it, but there is no way to determine that any frame of reference gives he "right" answer; nobody can physically determine if they are moving or at rest, if the motion is inertial. That'd be because there is no 'preferred' frame of reference.
asprung Posted April 6, 2008 Author Posted April 6, 2008 If he were facing in the direction of travel would he not become flat no matter what the frame of refrence. His width would shorten while his other measurements should not. By "width" I meant from his front to back.
insane_alien Posted April 7, 2008 Posted April 7, 2008 it would depend on the reference frame. if you were in a frame where he was travelling at zero velocity then he would appear normally proportioned, if you were observing from a frame where he was going 99.99999% of c then he would appear 'flat'. like has been said before, it all depends on the reference frame, there is no absolute reference frame which you seem to be implying.
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