Special theory of relativity question

[PAllen]

The principle of time dilation and length contraction are equivalent to lorentz transformation. Thus, applying time dilation and length contraction will give same result. I have applied the lorentz transformation result and don't see any difference. If you think I have done some mistake, please post your calculation here.

No, the Lorentz transformation also includes the features of relativity of simultaneity not captured by length contraction and time dilation.

The equations (for a vertical light path):

y=ct, x=0

becomes in the transformed coordinates (I will use x`, y`, t`):

y`/gamma - (v/c) x` = c t` , x` + v t` = 0

describing an angled light path. Despite being angled, the light goes from (x`,y`)=(0,0) at t`=0 to (-v, c/gamma) at t` = 1. Using euclidean distance formula on the x`,y` difference, you get c, so the speed is c/1 = c. Angled light, same speed.

Since even at a glance, it is obvious the Lorenz transform will not transform y=ct to y' = ct`, are you sure you don't want to change your pseudonym?

[EDIT: after a little more rearrangement, you get:

y` = ct`/gamma
x`= -vt`

]

 

[lovetruth]

No, the Lorentz transformation also includes the features of relativity of simultaneity not captured by length contraction and time dilation.

The equations (for a vertical light path):

y=ct, x=0

becomes in the transformed coordinates (I will use x`, y`, t`):

y`/gamma - (v/c) x` = c t` , x` + v t` = 0

describing an angled light path. Despite being angled, the light goes from (x`,y`)=(0,0) at t`=0 to (-v, c/gamma) at t` = 1. Using euclidean distance formula on the x`,y` difference, you get c, so the speed is c/1 = c. Angled light, same speed.

Since even at a glance, it is obvious the Lorenz transform will not transform y=ct to y' = ct`, are you sure you don't want to change your pseudonym?

You have done it all wrong.
y'=y=ct
x'=gamma*(x-vt)
t'=gamma*(t-vx/c^2)

 

[PAllen]

As I said I demand a formula giving the change in angle as a function of velocity v.

If a light ray has angle A in a given frame, than in a frame moving at speed v in the +x direction, the angle in the new frame (A`) is given by:

cot(A`) = (cot(A) - (v/c) cosec(A)) gamma

This differs from Galilean aberration by factor of gamma. I don't know if this effect has been observed - the difference from the Galilean formula is *extremely* small for the Earth's motion.

 

[PAllen]

You have done it all wrong.
y'=y=ct
x'=gamma*(x-vt)
t'=gamma*(t-vx/c^2)

This borders on insanity. You start with:

y=ct, x=0

to describe a light ray going in +y direction at x=0.
Then you make the following substitutions:

y' for y
(t'+x'v/c^2) gamma for t
(x' + v t') gamma for x

Rearrange, and you get what I wrote.

 

[lovetruth]

This borders on insanity. You start with:

y=ct, x=0

to describe a light ray going in +y direction at x=0.
Then you make the following substitutions:

y' for y
(t'+x'v/c^2) gamma for t
(x' + v t') gamma for x

Rearrange, and you get what I wrote.

check,
http://en.wikipedia.org/wiki/Lorentz_transformation

 

[PAllen]

check,
http://en.wikipedia.org/wiki/Lorentz_transformation

You don't know the first thing about using it. You are using it in the wrong direction, and you are using it wrong. To go from an equation in one frame to another, you need to substitute for all instances of x,y,t. Further, to go from an equation in x,y,t to x',y',t' you want to use the expressions for x in terms of (x',t'), t in terms of (t',x') etc.

 

[Mentz114]

All in all, I ask does any theory (including the relativity) predicts the light will move at an angle with respect to the light source when the light source is moving.

This is a good question so I'll have another go.

The answer is to use the wave model of light. When the light beam is turned on, imagine a circular wave front expanding from the emitter. After the moment of emission, the receiver moves a certain distance before the wavefront intersects with it. The light then appears to have traveled at an inclined path in the ground frame. For a collimated beam, a detailed analysis would show that interference only supports the path between the emitter and the detector it was aimed at in the rest frame.

So, the theory(s) required is wave optics, or QED as described by R. Feynman, with the assumption that speed of light is invariant.

 

[PAllen]

A couple of final points:

1) I gave the equation for a light path in (x`,y`,t`) given y=ct,x=0 in (x,y,t) as:

y` = ct`/gamma
x`= -vt`

Equivalently, if you have a light path in (x`,y`,t`) given by y`=ct`,x`=0 , you would have
in (x,y,t):

y = ct /gamma
x = vt

2) To amplify a little more on Mentz114 explanation:

Imagine observer B (moving at +v in x relative to A) sees his light source emit a single spherical wave front. Restricting to a plane, we have a circular wave front. B sees the wave front arrive simultaneously at both ends of his mirror, which he interprets as an orthogonal light signal, because his mirror is perpendicular to the line from the source to the center of the mirror.

A sees the light pulse with a spherical wave front in *his* frame. He sees B's mirror intersect the wave front at an angle (because it has moved since the signal was emitted). However, A sees the wave front arrive one end of B's mirror before the other end. A sees B calling these two separate arrival times simultaneous. Thus A sees how B interprets the signal as orthogonal, even though to A it is clearly received at an angle by B. Thus the explanation of the different perceived angle is the relativity of simultaneity.

 

[nitsuj]

This is a good question so I'll have another go.

The answer is to use the wave model of light. When the light beam is turned on, imagine a circular wave front expanding from the emitter. After the moment of emission, the receiver moves a certain distance before the wavefront intersects with it. The light then appears to have traveled at an inclined path in the ground frame. For a collimated beam, a detailed analysis would show that interference only supports the path between the emitter and the detector it was aimed at in the rest frame.

So, the theory(s) required is wave optics, or QED as described by R. Feynman, with the assumption that speed of light is invariant.

Wow, I cannot imagine the light clock with the light "beam/particle" as a wave. That's beyond me. (I thought light was like ping pong balls lol j/k)

 

[harrylin]

I think that since the speed of light isn't affected by the light source in any way, so must be the direction of the light.

That is wrong. Funny enough you phrased it correctly in your originally post: Relativity theory tells that the speed of light is constant in any inertial frame.

Think if instead of the laser, i have a point light source. Also, bullet and the gun analogy can't be applied here bcoz bullet is matter and light is light.

Both have momentum, and the law of conservation of momentum (unchanged since Newton!) applies to both

Both behave differently otherwise, we won't need theory of relativity in the first place. Otherwise, Galilean relativity should suffice then.
Thx for your valuable time thou

You're welcome.
What is different is, again, that SR has a limit speed, which is equal to the speed of light. I hope that it is clear now.

 

[PAllen]

If a light ray has angle A in a given frame, than in a frame moving at speed v in the +x direction, the angle in the new frame (A`) is given by:

cot(A`) = (cot(A) - (v/c) cosec(A)) gamma

This differs from Galilean aberration by factor of gamma. I don't know if this effect has been observed - the difference from the Galilean formula is *extremely* small for the Earth's motion.

While I don't know that the relativistic correction is big enough to see for effects of the Earth's changing velocity relative to astronomic sources, there is another context where the relativistic corrections is believed to be observed. This same equation explains (a portion of) relativistic beaming, where a larger solid angle of emitted light transforms to a smaller solid angle when the source is rapidly approaching; and the converse for a receding source. This is believed to explain why relativistically moving plasma jets from compact sources typically look similar if they are orthogonal to our line of view, but an approaching jet is *much* brighter than a receding jet.

 

[lovetruth]

This is a good question so I'll have another go.

The answer is to use the wave model of light. When the light beam is turned on, imagine a circular wave front expanding from the emitter. After the moment of emission, the receiver moves a certain distance before the wavefront intersects with it. The light then appears to have traveled at an inclined path in the ground frame. For a collimated beam, a detailed analysis would show that interference only supports the path between the emitter and the detector it was aimed at in the rest frame.

So, the theory(s) required is wave optics, or QED as described by R. Feynman, with the assumption that speed of light is invariant.

Thanx Mentz. You have clearly resolved all doubts. Light in nature is produced as spherical wave by oscillating charged particles. For gettin a linear beam of light, the 3D spherical wave is passed thru an aperture in some opaque screen.
If we imagine the linear light beam source is composed of a point light source and a opaque screen with an aperture, we can see the light beam comin out of aperture as a light beam coming out at an angle. If there is some distance between the point light source and the aperture, there will be a delay for the light to move from the point source to the movin aperture. Thus only an angled light beam will come out of aperture as calculated by simple geometry.
Please post ur comment if u think my understandin is wrong or u have a better answer.

 

[Mentz114]

Thanx Mentz. You have clearly resolved all doubts. Light in nature is produced as spherical wave by oscillating charged particles. For gettin a linear beam of light, the 3D spherical wave is passed thru an aperture in some opaque screen.
If we imagine the linear light beam source is composed of a point light source and a opaque screen with an aperture, we can see the light beam comin out of aperture as a light beam coming out at an angle. If there is some distance between the point light source and the aperture, there will be a delay for the light to move from the point source to the movin aperture. Thus only an angled light beam will come out of aperture as calculated by simple geometry.
Please post ur comment if u think my understandin is wrong or u have a better answer.

Yes, you are right about the aperture. It will move relative to the point source after the light is emitted and so produce an angled beam. No QED required.

I made a primitive movie of the situation with a point source. It's here

www.blatword.co.uk/space-time/wavemove.mpeg

size is only 560Kb.

 

[harrylin]

Yes, you are right about the aperture. It will move relative to the point source after the light is emitted and so produce an angled beam. No QED required.

Exactly, that's also what I meant with "if it went straight up then it would go through the side of the laser!"

I made a primitive movie of the situation with a point source. It's here

www.blatword.co.uk/space-time/wavemove.mpeg

size is only 560Kb.

Looks good!

Cheers,
Harald

 

[lovetruth]

Yes, you are right about the aperture. It will move relative to the point source after the light is emitted and so produce an angled beam. No QED required.

I made a primitive movie of the situation with a point source. It's here

www.blatword.co.uk/space-time/wavemove.mpeg

size is only 560Kb.

Thx for the reply.