see the attachment

referring to the pic 1 simultaneity was explained to me like this

there is a moving box car. At point A there is an observer in the boxcar at rest wrt the box car his name is Dash.

There is another observer not at rest wrt the box car at point C his name is Still Bill.

Lightening strikes both ends of the boxcar at point X and Z at time To.

Point C and A are equidistant from the ends of the boxcar say a distance L

Time for photons to reach point C from point X and Z is L/c so Still Bill thinks lightening strikes are simultaneous

As the boxcar is moving Dash, who was originally at point A, will some time later, say T1, be located at point B.

It is assumed Dash is moving relative to the photons coming from the lightening strikes.

Dash will see the photons from lightening strike Z before photons from lightening strike X so Dash will not think strikes are simultaneous. Simple!

Can I emphasise that Dash is assumed to be moving relative to the photons from the lightening strikes.

now look at picture 2
again moving boxcar
person in the box car at rest wrt the box car standing at point A again his name is Dash
lightening strikes at To at point A. ie the lightening strike is at the feet of Dash.
Again at T1 Dash, originally at point A, will be located at point B.
The photons from the lightening strike will move out in a perfect circle. The concentric circles show the photon front moving out from the lightening strike at various times.

I have not included the observer not at rest wrt the boxcar, Still Bill, as I don’t care what he sees.

Theory says Dash should see a sphere of photons moving away from him in all direction, clearly this is not the case. Dash was originally located at point A . At T1 he will be located at point B. If Dash at T1 when located at point B was to take a tape measure and measure the distance to the wave front of photons clearly he would get different measurements and therefore assume he is NOT in a sphere of photons?

Still referring to picture 2 The distance Dash measures to the photon front using his tape measure will not be a function of the velocity of the boxcar it will be a function of his absolute velocity.

So if you are in a moving frame of reference at rest wrt a point light source and you turn the light source on at say T0. If after a short delay, at say time T1, you measure the distance to the wave front from the point source of light.

If the distance to the wave front is the same in all directions then you can conclude you are absolutely stationary. As you are absolutely stationary the time on your clock is absolute time and the distance you measure is absolute distance.

for some reason the upload file thing is not working ? Keeps giving upload error To get the attachment please email me, probably some firewall port microsoft goddam thing

Pictures are below

Picture 1
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X ____________A_____B______Z

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C

Picture 2
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Imagine there are concentric circles radiating out from point A and Point A is at the centre of the circles

sorry for the crudeness of the pictures but the upload thingy wouldnt work for me. If you want non crude pics then email me

sorry all when I posted the reply the pictures got completely destroyed as it must remove multiple white space

robphy
Homework Helper
Gold Member
sorry all when I posted the reply the pictures got completely destroyed as it must remove multiple white space

Try the [CODE ] [/CODE ] tags... no internal spaces in the tags.

Code:
 A  B  C           D

Dale
Mentor
2021 Award
We already went over this in your light bulb thread. It is a simple matter to show that a spherical wavefront propagating at c in one frame is a spherical wavefront propagating at c in all frames. Just take the Lorentz transform.

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Picture 1
__________________________
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| |
| |
| |
| 0 |
| -|- |
| /\ |
X ____________A_____B______Z

0
-|-
/\
C

Picture 2
__________________________
| |
| |
| |
| |
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| 0 |
| -|- |
| /\ |
|_____________A______B_____|

Imagine there are concentric circles radiating out from point A and Point A is at the centre of the circles

with quotes ~

nope quotes didnt work either

Can I ask a big favour can someone email me

JUJUYUMYUM@gmail.com

We already went over this in your light bulb thread. It is a simple matter to show that a spherical wavefront propagating at c in one frame is a spherical wavefront propagating at c in all frames. Just take the Lorentz transform.

I agree with you photons do radiate in a sphere from a point light source in all frames.

If you have an observer who is moving and who is at rest wrt a point light source, which obviously is also moving. When the light source is turned on what I am saying is that if the observer sees a sphere of photons then the explanation of simultenaity is wrong and on the flip side if the explanation of simultesity is right then the observer will NOT see a sphere of photons...which is it? it cant be both!

See the pictures before you reply. I may be wrong but I cant see the error in my logic given the way simultenaiety is explained.

If the distance to the wave front is the same in all directions then you can conclude you are absolutely stationary. As you are absolutely stationary the time on your clock is absolute time and the distance you measure is absolute distance.
You've misunderstood the import of relativity if you think that there is such a thing as absolute velocity. How can anything be absolutely stationary if one can always find a frame in which it moves ?

Please go away and learn relativity properly.

Hi Mentz

have you seen the pics?

I know it is not intuitive but I cant see where my logic is wrong?

In my post is the explanation of simulteniaty wrong? or the other way round?

yes, but I couldn't relate to them. Simultaneity is observer dependent. If there are two space separated events at which light is emitted in all directions, all observers who are equi-distant ( or equi-time) away from the those events will say they are simultaneous, all others will disagree. There's not much more to be said and no potential paradoxes lurking.

M

if you have a sphere of photons and assuming photons travel in a striaght line out from a central point at a constant velocity C ie they form a sphere.

Remebering photons are not ballistic ie the velocity of the frame from where the photons originate does NOT add to the velocity of the photons.

Wont the point that represents the centre of that sphere always be at the same point in space? if not why not?

so if you found a sphere of photons and if you measured back from the wave front of that sphere and found the centre of that sphere and then flew a space ship in the right direction and speed so that you always remained at the centre of that sphere then the space ship would be at the same point in space at all times. Evereything would probaby be moving around you but you would be theoretically stationary

or am I worng?

so referring to picture 2

will the observer measure the same distance to the wavefront in all directions?

so if you found a sphere of photons and if you measured back from the wave front of that sphere and found the centre of that sphere and then flew a space ship in the right direction and speed so that you always remained at the centre of that sphere then the space ship would be at the same point in space at all times. Everything would probaby be moving around you but you would be theoretically stationary
You would be stationary in the frame of the matter that emitted the sphere of light. The matter might not be there anymore ( as with the CMBR ).

I think you just invented the auto-pilot.

with post #14 I am ignoring the effects of gravity as gravity would distort a sphere of photons so that it was no longer a sphere

I think I see where the misunderstading is coming from. If the light source is on for a period of time photons are continually streaming from the light soruce and so the centre of that stream is not stationary it will form a cone shape

I read in a science journal that there is a single photon light source now.

So if the single photons light source is in a moving frame of ref and it emitts a very brief flash of light so you have a sphere of single photons radiating out from the matter that emitted the photons in all directions. The centre of that sphere of photons is a stationary point in space for all time, ignoring gravity

Mentz

referring to picture 2 of my post
changing the wording to the light source emits a sphere of photons one photon thick in all directions

will the observer measure the same distance to the wavefront in all directions or not?

I cant figure this out in my head I see a paradox

I forgot to say in my post that the flash of light is a very brief flash, so brief that only a single photon is relased in all directions.

So if the single photons light source is in a moving frame of ref and it emitts a very brief flash of light so you have a sphere of single photons radiating out from the matter that emitted the photons in all directions. The centre of that sphere of photons is a stationary point in space for all time, ignoring gravity

It's an identifiable point in space, but it is not 'stationary'. There's no such thing.

Suppose I observe a sphere of light emitted from a source that was moving relative to me. Now your stationary point is moving.

There no such thing as 'stationary' unless you say in what frame it is stationary.

Actually thinking more about it the flash doesnt have to be brief

Mentz referring to picture 2 of my post will the observer at T0 measure the same distance to the wavefront in all directions ?

Dale
Mentor
2021 Award
When the light source is turned on what I am saying is that if the observer sees a sphere of photons then the explanation of simultenaity is wrong and on the flip side if the explanation of simultesity is right then the observer will NOT see a sphere of photons...which is it? it cant be both!
It can be both. Both the relativity of simultaneity and the fact that a spherical wavefront propagating at c in one frame is a spherical wavefront propagating at c in all frames are direct results of the Lorentz transforms. In fact, the spherical wavefront part is a restatement of the second postulate which (together with the first postulate) implies the relativity of simultaneity.

Referring to picture 2 and picture 1 of post #1

picture 1
you have the explanation of simultenaity from Serway (text) where the observer in the MFR is moving relative to the photon wave front

pic 2

Dash is at rest wrt a MFR when lightening strikes at Dash's feet.

At time T0 Dash will be located at A. At T1, some time later, he will be at point B.

The photon wave front from the lightening will form a perfect sphere about point A.

Will Dash's measurements be the same in all directions or have I made some other fundamental error?

JesseM
see the attachment

referring to the pic 1 simultaneity was explained to me like this

there is a moving box car. At point A there is an observer in the boxcar at rest wrt the box car his name is Dash.

There is another observer not at rest wrt the box car at point C his name is Still Bill.

Lightening strikes both ends of the boxcar at point X and Z at time To.
Presumably you mean "at time T0" in Still Bill's frame, since in Dash's frame the lightning strikes did not happen simultaneously so they would not have the same time-coordinate.
Point C and A are equidistant from the ends of the boxcar say a distance L

Time for photons to reach point C from point X and Z is L/c so Still Bill thinks lightening strikes are simultaneous

As the boxcar is moving Dash, who was originally at point A, will some time later, say T1, be located at point B.

It is assumed Dash is moving relative to the photons coming from the lightening strikes.

Dash will see the photons from lightening strike Z before photons from lightening strike X so Dash will not think strikes are simultaneous. Simple!
Yes, that's right.
Can I emphasise that Dash is assumed to be moving relative to the photons from the lightening strikes.
What does that mean? The photons are moving at c in both observer's own rest frame. Maybe you just mean that in Still Bill's rest frame, Dash is moving towards the position of strike Z and away from the position of strike X, so in Still Bill's frame the distance between Dash and the photons from Z is growing smaller faster than the distance between Dash and the photons from X, which explains why the photons from Z hit Dash first in Still Bill's frame. In Dash's frame, though, Dash is at rest, and the distance between himself and the photons from each strike is closing at the same rate, namely c. In Dash's frame the only reason the photons from Z hit him first is that the strike at Z happened at an earlier time than the strike at X.
now look at picture 2
again moving boxcar
person in the box car at rest wrt the box car standing at point A again his name is Dash
lightening strikes at To at point A. ie the lightening strike is at the feet of Dash.
Again at T1 Dash, originally at point A, will be located at point B.
The photons from the lightening strike will move out in a perfect circle. The concentric circles show the photon front moving out from the lightening strike at various times.

I have not included the observer not at rest wrt the boxcar, Still Bill, as I don’t care what he sees.

Theory says Dash should see a sphere of photons moving away from him in all direction, clearly this is not the case. Dash was originally located at point A . At T1 he will be located at point B. If Dash at T1 when located at point B was to take a tape measure and measure the distance to the wave front of photons clearly he would get different measurements and therefore assume he is NOT in a sphere of photons?
No, your picture 2 only shows how things look in the frame where the box car is moving, namely Still Bill's frame. In Still Bill's frame Dash is closer to one side of the sphere than the other. However, in Dash's own frame he is equidistant from both sides of the sphere.

Try it with some numbers. Assume Still Bill's frame uses unprimed coordinates x and t, while Dash's frame uses primed coordinates x' and t'. Say that in Still Bill's frame, Dash is moving at 0.6c in the +x direction, and he's at position x=0 when the strike hits near his feet at t=0. Also suppose that Dash has two clocks on board the train which are equidistant from him, one 10 light-seconds to his left and one 10 light-seconds to his right in Dash's own frame, and that these clocks are synchronized in Dash's frame. In Still Bill's frame things are a little different--because of the Lorentz contraction factor of $$\sqrt{1 - 0.6^2}$$ = 0.8, each clock is only 8 meters away from Dash, so at t=0 one clock is at x=-8 and the other clock is at x=8 in Still Bill's frame. What's more, if two clocks are a distance of x apart and synchronized in their own rest frame, then in a frame where they're moving at speed v they will be out-of-sync by vx/c^2. So since the two clocks are 20 light-seconds apart and synchronized in Dash's frame, that means in Still Bill's frame they are out-of-sync by (0.6c)*(20 light-seconds)/c^2 = 12 seconds. If Dash's clock reads t'=0 at the moment the lightning strikes at his feet, this means that in Still Bill's frame at t=0 the clock at x=-8 shows a time of 6 seconds, while the clock at x=8 shows a time of -6 seconds (12 seconds apart).

Now if the light expands in a sphere in Bill's frame, we can figure out when the light will strike each of Dash's two clocks. At t=0 the left clock is at x=-8, and it's moving at 0.6c, so 5 seconds later it'll be at x=-8 + 0.6*5 = -8 + 3 = -5. And of course if the light starts out at x=0, 5 seconds later the light going in the -x direction will also be at x=-5. So, at t=5 in Still Bill's frame, the light hits the left of Dash's clocks. How about the right clock? Well, at t=0 the right clock is at x=8, and it's moving at 0.6c, so 20 seconds later it'll be at x=8 + 0.6*20 = 20. And since the light starts out at x=0, 20 seconds later the light going in the +x direction will also be at x=20. So at t=20 in Still Bill's frame, the light hits the right of Dash's two clocks.

But still looking at things from the perspective of Still Bill's frame, we can figure out what Dash's clocks will read at each of these moments. Remember, at t=0 the left clock reads 6 seconds and the right clock reads -6 seconds in Bill's frame. And since both these clocks are moving at 0.6c, they are both slowed down in their rate of ticking by 0.8 in Bill's frame. So at t=5 both clocks have only ticked 0.8*5=4 seconds forward, meaning the left clock will read 6 + 4 = 10 seconds when the light reaches it (the right clock will read -6 + 4 = 2 seconds at this moment, but that's not really important). And at t=20 both clocks have ticked forward by 0.8*20=16 seconds, so the right clock reads -6 + 16 = 10 seconds when the light strikes it as well. So you can see that analyzing everything from the perspective of Still Bill's frame, and assuming the light moves at the same speed in both directions in Still Bill's frame, we end up concluding that both of Dash's clocks read 10 seconds when the light hits them, even though it hit each clock at a different time in Still Bill's frame. Of course these local facts about what the clocks read when the light hit them must still be true in Dash's own frame, but in Dash's frame the two clocks were actually synchronized, so the light must have struck both clocks at the same time t'=10 in Dash's frame. Since the clocks are also both at rest in Dash's frame and equidistant from where the lightning struck at Dash's feet, this shows that in Dash's frame he actually does remain at the center of the expanding light sphere.

Dale
Mentor
2021 Award
have I made some other fundamental error?
your fundamental error is in not using the Lorentz transform.