Is this for real?

[Robittybob1]

 

This is incomprehensible, so it is not answerable.

@xyzt - You found what I wrote a little difficult to understand so I'll have another go, but that was an answer to your question, you asked: "Who said that they are accelerating?". So I gave you a rough answer but I will expand it.

"To the remote observer the light bouncing across the light clock will appear be also falling and hence it is traveling further therefore the frame's time is dilated (the ticks of the falling clock will be getting longer by the tick, the time dilation will increase as the free-fall continues.)

From the point of view of the local observer that time dilation will keep the "velocity locally constant" The light will travel further in a longer period of time which is what you need to get a constant velocity of light.

 

From the frame of the local observer they are not accelerating - I understand that.

 

 

The light in the light clock will for an outside observer be falling and hence traveling further hence the time is dilated (the ticks of the falling clock will be getting longer by the tick.) Does that keep their "velocity locally constant" Further in a longer period of time sounds like the beginning of constant velocity.

From the frame of the local observer they are not accelerating - I understand that.

Now did that help?

Edited January 7, 2015 by Robittybob1

[Theoretical]

One piece of advice: you will NEVER learn physics via (selective) quoting.

 

The top two sites I cited were:

physics.stackexchange.com physicsforums.com

 

Truth is you knew exactly what my question was. In fact the physicists at physicsforums.com even used the same word I used to describe it. But you didn't know the correct answer. So your big ego tried to turn it on me. Save your words. I'm not interested in what you say.

[elfmotat]

Let's try again, using math instead of psychobabble:

 

Instead of trying to argue and hurl petty insults, you may find my post enlightening if you actually read it. I was attempting to end the argument between yourself and MigL by explaining where the disagreement is coming from. To first order light and massive particles will fall identically. Globally they won't. MigL was considering the situation locally and you were considering it globally. That's where the disagreement was coming from, and you're both right in a way.

 

[math]0 =- \Gamma^{a}_{bc}\frac{dx^b}{ds}\frac{dx^c}{ds} [/math]

 

For massive particles, the left hand side of the equation is NOT zero. The right hand side is the same, for both photons and for massive particles."

 

For someone so intent on making sure others "understand the math," the irony here is palpable. This is just completely false. This is not even a generally covariant equation; it is completely coordinate-dependent. I don't know where you got it from, but it doesn't even make sense. I believe you're mixing the geodesic equation up with the null-condition for massless particles.

Edited January 7, 2015 by elfmotat

[MigL]

I understand the math just fine xyzt, just don't do LaTex.

You don't seem to understand English and seem to have a bad disposition.

 

You are right in your remark to Theoretical that photons cannot accelerate though.

[Theoretical]

I understand the math just fine xyzt, just don't do LaTex.

You don't seem to understand English and seem to have a bad disposition.

 

You are right in your remark to Theoretical that photons cannot accelerate though.

I agree about him, but I never said photons accelerate. I said that I didn't know the correct wording and to replace accelerate with the correct word.

Just wanted to clarify.

[xyzt]

To first order light and massive particles will fall identically.

"Fall" means nothing but psychobabble. It is very simple, really, as I tried to explain to you earlier (sorry I failed). One starts with the metric (Schwarzschild, for example):

 

[math]ds^2=(1-r_s/r)(cdt)^2-dr^2/(1-r_s/r)-r^2d\phi^2[/math]

 

Since you had so much difficulty with understanding the geodesics equations, I'll explain another way, via the Euler-Lagrange formalism. Divide by [math]ds^2[/math] and you get the Lagrangian:

 

[math]L=(1-r_s/r)(cdt/ds)^2-(dr/ds)^2/(1-r_s/r)-r^2(d\phi/ds)^2[/math]

 

From the above, one gets the Euler-Lagrange EQUATIONS of MOTION (fully equivalent to the equations I presented earlier, in terms of Christoffel symbols):

 

[math](1-r_s/r)dt/ds=k[/math]

[math]r^2 d\phi/ds=h[/math]

 

Solve the above and you get the trajectory for a massive particle. The above is the same thing (written differently) as my earlier [math]\frac{d^2x^a}{ds^2} =- \Gamma^{a}_{bc}\frac{dx^b}{ds}\frac{dx^c}{ds} [/math],

 

For the photon, it is a lot simpler, start with :

 

[math]0=ds^2=(1-r_s/r)(cdt)^2-dr^2/(1-r_s/r)-r^2d\phi^2[/math]

 

The above produces:

 

[math](1-r_s/r)dt/d \lambda=\bar{k}[/math]

[math]r^2d\phi/d \lambda=\bar{h}[/math]

[math](dr/d\ \lambda)^2=\bar{k}^2-(1-r_s/r)(\bar{h}/\bar{r})^2[/math]

 

where [math]\lambda[/math] is the affine parameter, different from ds. The above is the same thing (written differently) as my earlier [math]0 =- \Gamma^{a}_{bc}\frac{dx^b}{d \lambda}\frac{dx^c}{d \lambda} [/math].

Solving the above gets the light bending in a gravitational field. Contrary to your repeated claims, the photons and the massive particles don't "fall the same way". whatever that might mean. They take different paths since their equations of motion are different. For example if you let go of a test probe and of a photon from the same point, they will follow different trajectories.

Edited January 7, 2015 by xyzt

[elfmotat]

"Fall" means nothing but psychobabble. It is very simple, really, as I tried to explain to you earlier (sorry I failed). One starts with the metric (Schwarzschild, for example):

 

[math]ds^2=(1-r_s/r)(cdt)^2-dr^2/(1-r_s/r)-r^2d\phi^2[/math]

 

Since you had so much difficulty with understanding the geodesics equations, I'll explain another way, via the Euler-Lagrange formalism. Divide by [math]ds^2[/math] and you get the Lagrangian:

 

[math]L=(1-r_s/r)(cdt/ds)^2-(dr/ds)^2/(1-r_s/r)-r^2(d\phi/ds)^2[/math]

 

From the above, one gets the Euler-Lagrange EQUATIONS of MOTION (fully equivalent to the equations I presented earlier, in terms of Christoffel symbols):

 

[math](1-r_s/r)dt/ds=k[/math]

[math]r^2(d\phi/ds)^2=h[/math]

 

Solve the above and you get the trajectory for a massive particle.

 

For the photon, it is a lot simpler, start with :

 

[math]0=ds^2=(1-r_s/r)(cdt)^2-dr^2/(1-r_s/r)-r^2d\phi^2[/math]

 

Solving the above gets the light bending in a gravitational field. Contrary to your repeated claims, the photons and the massive particles don't "fall the same way". whatever that might mean. They take different paths since their equations of motion are different.

 

"Free-fall" is by definition geodesic motion. I assumed this was understood, because it commonly is. You have quite a nasty attitude, which makes it difficult to talk to you. I was trying to explain something, and you chose to ignore it. "First order" is also a well-defined term which you have, out of ignorance, labeled as "psychobabble." You also ignored the part where I called you out for using a nonsensical equation. I'm well aware of the everything else you just took the unnecessary time write, and have been for quite a while. The sheer condescension is making your posts a bit unbearable to read, so I recommend toning that down a bit as well.

Edited January 7, 2015 by elfmotat

[xyzt]

 

"Free-fall" is by definition geodesic motion.

 

Yes, but I was objecting to something ELSE: your use of "fall the same way".

 

 

 

I assumed this was understood, because it commonly is. You have quite a nasty attitude, which makes it difficult to talk to you. I was trying to explain something, and you chose to ignore it. "First order" is also a well-defined term which you have, out of ignorance, labeled as "psychobabble."uite a while. The sheer condescension is making your posts a bit unbearable to read, so I recommend toning that down a bit as well.

What I labelled as psychobabble is your prose that reflects the lack of understanding of the basic equations of motion, as derived for massive vs. massless particles.

 

 

 

You also ignored the part where I called you out for using a nonsensical equation. I'm well aware of the everything else you just took the unnecessary time write, and have been for q

 

I did not ignore it at all, I actually answered. The fact that you cannot follow the answer is your problem.

 

"For the photon, it is a lot simpler, start with :

 

[math]0=ds^2=(1-r_s/r)(cdt)^2-dr^2/(1-r_s/r)-r^2d\phi^2[/math]

 

The above produces:

 

[math](1-r_s/r)dt/d \lambda=\bar{k}[/math]

[math]r^2d\phi/d \lambda=\bar{h}[/math]

[math](dr/d\ \lambda)^2=\bar{k}^2-(1-r_s/r)(\bar{h}/r)^2[/math]

 

where [math]\lambda[/math] is the affine parameter, different from ds. The above is the same thing (written differently) as my earlier [math]0 =- \Gamma^{a}_{bc}\frac{dx^b}{d \lambda}\frac{dx^c}{d \lambda} [/math].

Solving the above gets the light bending in a gravitational field. Contrary to your repeated claims, the photons and the massive particles don't "fall the same way". whatever that might mean. They take different paths since their equations of motion are different. For example if you let go of a test probe and of a photon from the same point, they will follow different trajectories."

 

The sheer condescension is making your posts a bit unbearable to read, so I recommend toning that down a bit as well.

 

I recommended earlier that you stopped trolling my posts. You keep imagining errors in my posts and you are rubbishing the thread.

Edited January 7, 2015 by xyzt

[Robittybob1]

 

"Free-fall" is by definition geodesic motion. I assumed this was understood, because it commonly is. You have quite a nasty attitude, which makes it difficult to talk to you. I was trying to explain something, and you chose to ignore it. "First order" is also a well-defined term which you have, out of ignorance, labeled as "psychobabble." You also ignored the part where I called you out for using a nonsensical equation. I'm well aware of the everything else you just took the unnecessary time write, and have been for quite a while. The sheer condescension is making your posts a bit unbearable to read, so I recommend toning that down a bit as well.

This article covers free-fall at a very basic level but doesn't get into time dilation.

http://www.einstein-online.info/spotlights/equivalence_principle

[elfmotat]

 

Yes, but I was objecting to something ELSE: your use of "fall the same way".

 

Okay, this time I'll be precise: for a short enough time over short enough distances, free-fall reference frames are equivalent to inertial reference frames. Which means that over small enough times/distances if you were to shine a laser at a detector to try to measure some light deflection, you won't be able to. Because it's locally an inertial frame.

 

What I labelled as psychobabble is your prose that reflects the lack of understanding of the basic equations of motion, as derived for massive vs. massless particles.

 

I can't argue with an argument with no content.

 

I did not ignore it at all, I actually answered. The fact that you cannot follow the answer is your problem.

I recommended earlier that you stopped trolling my posts. You keep imagining errors in my posts and you are rubbishing the thread.

 

Are you really going to pretend you don't know how the forum works? I pressed "quote" a few minutes before I responded. I responded a minute after you edited that post, meaning I had no knowledge of your edit at the time I posted. Now you're going to pretend I'm intentionally ignoring things? Very dishonest.

 

 

The above is the same thing (written differently) as my earlier [math]0 =- \Gamma^{a}_{bc}\frac{dx^b}{d \lambda}\frac{dx^c}{d \lambda} [/math].

Solving the above gets the light bending in a gravitational field. Contrary to your repeated claims, the photons and the massive particles don't "fall the same way". whatever that might mean. They take different paths since their equations of motion are different. For example if you let go of a test probe and of a photon from the same point, they will follow different trajectories."

 

Please derive the equation [math]\Gamma^{\mu}_{\nu \sigma}\frac{dx^\nu}{d \lambda}\frac{dx^\sigma}{d \lambda} =0[/math] from the following:

• [math]ds^2=0[/math]
• [math]\frac{d^2 x^{\mu}}{d \lambda^2} + \Gamma^{\mu}_{\nu \sigma}\frac{dx^\nu}{d \lambda}\frac{dx^\sigma}{d \lambda}=0[/math]

 

I'll give you a hint: it can't be done. Because it's nonsense. That equation is meaningless. It's coordinate-dependent, which means it's not generally covariant, so it means absolutely nothing in the context of GR.

Edited January 7, 2015 by elfmotat

[xyzt]

 

Okay, this time I'll be precise: for a short enough time over short enough distances, free-fall reference frames are equivalent to inertial reference frames. Which means that over small enough times/distances if you were to shine a laser at a detector to try to measure some light deflection, you won't be able to. Because it's locally an inertial frame.

Yes, correct. But this is not the subject of our disagreement.

 

 

 

I can't argue with an argument with no content.

 

This IS the subject of our disagreement. In simple( r ) words, since you have so much trouble understanding: the math shows that massive particles follow a DIFFERENT trajectory than massless particles. In your parlay, the do NOT "fall the same way". Do you get this?

Edited January 7, 2015 by xyzt

[elfmotat]

Yes, correct. But this is not the subject of our disagreement.

 

It was the subject your and MigL's disagreement, which is why I brought it up.

 

This IS the subject of our disagreement. In simple( r ) words, since you have so much trouble understanding: the math shows that massive particles follow a DIFFERENT trajectory than massless particles. In your parlay, the do NOT "fall the same way". Do you get this?

 

Ignoring the part about your nonsense equation, I completely agree and I never said differently. If you had been less reactionary from the get go this would have been obvious. I said they "fall the same way" over small enough times/distances.

Edited January 7, 2015 by elfmotat

[xyzt]

 

It was the subject your and MigL's disagreement, which is why I brought it up.

No, it wasn't, the subject was his false claims about my post due to his misrepresentation of my post.

 

 

 

 

 

If you had been less reactionary from the get go this would have been obvious. I said they "fall the same way" over small enough times/distances.

 

This is probably the closest you'll come to admitting that you were wrong all along.

Edited January 7, 2015 by xyzt

[elfmotat]

No, it wasn't, the subject was his false claims about my post due to his misunderstanding.

 

This is probably the closest you'll come to admitting that you were wrong all along.

 

Except that was the cause of your disagreement, and I wasn't wrong. Nothing I've said so far is false. On the other hand, you've given us a meaningless equation and barrels of insults.

Edited January 7, 2015 by elfmotat

[xyzt]

 

Except that was the cause of your disagreement, and I wasn't wrong. Nothing I've said so far is false. On the other hand, you've given us a meaningless equation and barrels of insults.

Actually, you are outright lying. Here is the proof. Do us a favor and stop rubbishing the thread.

Edited January 7, 2015 by xyzt

[Robittybob1]

 

This IS the subject of our disagreement. In simple( r ) words, since you have so much trouble understanding: the math shows that massive particles follow a DIFFERENT trajectory than massless particles. In your parlay, they do NOT "fall the same way". Do you get this?

I have feeling you are still disagreeing that photons fall as fast as lead balls. One is a mass-less particle and the other massive. I appreciate they will take different trajectories but will the downward component due to gravity be the same in the same period of time. OK the time period has to be infinitesimal for the photon moves off at the speed of light?

Edited January 7, 2015 by Robittybob1

[elfmotat]

Actually, you are outright lying.

 

Well, since there's apparently no content left in this discussion, I think I'll be on my way.

[xyzt]

I have feeling you are still disagreeing that photons fall as fast as lead balls. One is a mass-less particle and the other massive. I appreciate they will take different trajectories but will the downward component due to gravity be the same in the same period of time. OK the time period has to be infinitesimal for it moves off at the speed of light?

Nope, you are as wrong as before. A photon and a lead ball will "fall" at DIFFERENT speeds towards a gravitating mass. This is true from the perspective of any observer. I already pointed out your error when judged from the POV of a freefalling observer. (light "falls" at c, the lead ball falls at v=0).

Edited January 7, 2015 by xyzt

[Robittybob1]

Nope, you are as wrong as before.

How would you compare how far light will fall due to gravity over 1 meter parallel to the surface of the Earth compared to how far a lead ball will fall in the same amount of time?

This is nearly the same question on the net.

https://van.physics.illinois.edu/qa/listing.php?id=2010

 

 

Q:

if a laser beam is shot horizontally will it curve downwards towards earth because of gravity? (like shooting a bullet out of a gun)

- owen (age 16)

chicago

A:

Owen -

Technically, the answer to this question is yes. But for all practical purposes, it's no. Let me explain...

For most practical purposes, light moves in a straight line. However, it does bend a little due to gravity. The small curvature of light passing near the Sun was first observed in 1919. The agreement of that curvature with the prediction of General Relativity was the first big piece of evidence for General Relativity. However, the curvature near the Earth is much smaller even than near the Sun, because the Earth is very small compared to the Sun. Really big curvature of light by gravity occurs near the strange objects called black holes, which were also predicted by General Relativity.

So yes, the light ray from your laser will technically be bent by the earth's gravity. But no, the effect will not be enough to be noticeable.

From this site http://www.einstein-online.info/spotlights/light_deflection

this quote:

Light deflection in general relativity

Over a century later, in the early 20th century, Einstein developed his theory of general relativity. Einstein calculated that the deflection predicted by his theory would be twice the Newtonian value.

 

What does that mean? Is light falling twice as fast as lead balls?

Edited January 7, 2015 by Robittybob1

[xyzt]

How would you compare how far light will fall due to gravity over 1 meter parallel to the surface of the Earth compared to how far a lead ball will fall in the same amount of time?

In the freefalling frame the lead ball doesn't fall, it floats. Did you get that? It is the third time I explained it to you. Neither does the light, it moves parallel to the "floor".

In an accelerated frame, like the observer hovering at infinity, the situation is different.

Edited January 7, 2015 by xyzt

[Robittybob1]

In the freefalling frame the lead ball doesn't fall, it floats. Did you get that? It is the third time I explained it to you. Neither does the light, it moves parallel to the "floor".

In an accelerated frame, like the observer hovering at infinity, the situation is different.

We are not dealing with free falling frames now but the surface of the Earth, with gravity at g and light traveling at c.

Edited January 7, 2015 by Robittybob1

[xyzt]

We are not dealing with free falling frames now but the surface of the Earth, with gravity at g and light traveling at c.

YOU, are not dealing with anything, let's make this clear.

In the Earth frame, being an accelerated frame, the paths of the light and the lead ball are different. You will never be able to learn physics by posting questions. You need to go to school.

[Robittybob1]

5.45755E-17 m this is how far a lead ball will fall in the time light travels 1 meter. The geodesic for the lead ball is vertical as it has no lateral motion.

If the light follows the space-time curvature of the Earth how much curvature would we see over 1 meter horizontal to the Earth surface?

[Strange]

If the light follows the space-time curvature of the Earth how much curvature would we see over 1 meter horizontal to the Earth surface?

 

I don't understand the point of this question. Unless you have a specific problem to solve, what do you expect to gain from this piece of information? It won't help you understand anything more general.

[MigL]

The original disagreement stems from Robbitybob's post #27 where he asks about a free falling light clock, a beam of light bouncing back and forth, from side to side, say in a free-falling elevator. He wants to know whether the light beam will show a deflection from the horizontal or not . To which xyzt replied that since they follow different trajectories, yes, there will be a deflection from the horizontal.

That, is what I had a problem with.

And I would imagine its now time to move on.

Edited January 7, 2015 by MigL