Is Relativity Broken Due to Simultaneity and Time Dilation?

[firavia]

Consider you (A) are moving at the speed of 180 000 km/s and another person (B) moving at an opposite direction to you at 100 000km/s. what would be the speed of the either relative to each other? Both would appear to be approaching each other at the speed of sum of both of their speeds. (i.e 280 000km/s).
Now consider “A” to be traveling at the speed of 180 000 km/s and “B” (light) traveling at the speed of 300 000 km/s. In this case B’s speed with reference to A is not 480 000km/s as presumed but it’s 300 000 km/s speed of light is the same in all( moving and still) Frame of reference.

But if this assumption is true which is than according to "A" in one second light traveled only 300 000 -180 000 km = 120 000 km , is this assumption true did light travel according to "A" 120000 km in 1 sec because relatevely to him the spped is 300 000 km/s ? please help

 

[Doc Al]

Consider you (A) are moving at the speed of 180 000 km/s and another person (B) moving at an opposite direction to you at 100 000km/s. what would be the speed of the either relative to each other? Both would appear to be approaching each other at the speed of sum of both of their speeds. (i.e 280 000km/s).

No, the relative speed must be obtained using the relativistic addition of velocity:

$$V_{a/c} = \frac{V_{a/b} + V_{b/c}}{1 + (V_{a/b} V_{b/c})/c^2}$$

Which gives you a relative speed of about 265 000 km/s.

Now consider “A” to be traveling at the speed of 180 000 km/s and “B” (light) traveling at the speed of 300 000 km/s. In this case B’s speed with reference to A is not 480 000km/s as presumed but it’s 300 000 km/s speed of light is the same in all( moving and still) Frame of reference.

Right. (The above formula is based on the fact that the speed of light is invariant.)

Don't confuse the relative speed of B with respect to A (the speed of B as seen by A) with the "closing speed" of A and B as seen by some third party. For example if A travels 180 000 km/s with respect to Earth, then Earth observers will see A and the light "close" at the rate of 300 000 + 180 000 = 480 000 km/s. But Earth observers will see the light move at speed 300 000 km/s, just like everyone else. (Note that no one sees anything moving at 480 000 km/s.)

But if this assumption is true which is than according to "A" in one second light traveled only 300 000 -180 000 km = 120 000 km , is this assumption true did light travel according to "A" 120000 km in 1 sec because relatevely to him the spped is 300 000 km/s ? please help

I don't understand where you get this. According to A, light travels 300 000 km in one second.

 

[firavia]

so u mean that according to "A" light has traveled only 300 000 km in 1 sec ?

lets take another example , if "A" is movig at a speed of 200 000 km/s and carry with it a flash light at t =1 s according to "A" the flash light was turned on , so according to "A" the speed of light is going to be 300 000 km/s which means at t=2s according to "A" light has traveled from point "1" at t= "1 s" to point "2" at t ="2s" only 300 000 km ? is that true and "A" has traveled 200 000 km and the location of the photons at t= "2s" is in advance of 100 000 km of the location of "A" at = 2 s ? pl;ease reply

 

[DaveC426913]

so u mean that according to "A" light has traveled only 300 000 km in 1 sec ?

lets take another example , if "A" is movig at a speed of 200 000 km/s and carry with it a flash light at t =1 s according to "A" the flash light was turned on , so according to "A" the speed of light is going to be 300 000 km/s which means at t=2s according to "A" light has traveled from point "1" at t= "1 s" to point "2" at t ="2s" only 300 000 km ? is that true and "A" has traveled 200 000 km and the location of the photons at t= "2s" is in advance of 100 000 km of the location of "A" at = 2 s ? pl;ease reply

Yes. From A's frame of reference, he measures the light traveling a distance of 300,000km in a time of one second.

An observer on Earth measures the light traveling 300,000km in a time of one second.

But that one second for A is not the same as the one second for Earth. So they are not agreeing on the simultaneity of the events.

If the Earth observer were able to peer inside A's spaceship, he would see A's stopwatch ticking slower. After one second, it will have appeared to tick only ~2/3 of a second. A will not click his stopwatch for another ~1/3 of a second.

 

[firavia]

please through the image I attached I would like to know ur opinon and to correct me if I am wrong somewhere thanks u ..
I am missing a point when a man is inside a spacecraft that is traveling at speed of 200 000 km and at a moment t = 1 s a flash light was tuned on in his spacecraft after 1 second from turning on his flash light >> "t = 2s " where would be the position of the photon , will it be 300 000 km away from the spaceship and if yes dosent it mean that light is at 500 000 km away from the position it had before at "t =1s because simply the spacecraft is moving at the same time of light movement at 200 000 km/s "

 

[Doc Al]

so u mean that according to "A" light has traveled only 300 000 km in 1 sec ?

Absolutely. This means that A, using his own clocks and metersticks, measures the light to travel 300 000 km in 1 second.

lets take another example , if "A" is movig at a speed of 200 000 km/s and carry with it a flash light at t =1 s according to "A" the flash light was turned on , so according to "A" the speed of light is going to be 300 000 km/s which means at t=2s according to "A" light has traveled from point "1" at t= "1 s" to point "2" at t ="2s" only 300 000 km ? is that true and "A" has traveled 200 000 km and the location of the photons at t= "2s" is in advance of 100 000 km of the location of "A" at = 2 s ? pl;ease reply

Here's how it works. Let's say that A is moving at a speed of 200 000 km/s with respect to Earth, and that A sets his clock to read t = 0 just as he passes Earth. He also turns on his flashlight at that same time, sending a beam of light out in the forward direction. In one second (according to A's clock) where is A and where is the beam of light? According to A: At t = 1 second, A is 200 000 km from Earth and the light beam has traveled 300 000 km from A. Thus A says the edge of the beam is 500 000 km from Earth.

Got it?

 

[Doc Al]

I am missing a point when a man is inside a spacecraft that is traveling at speed of 200 000 km and at a moment t = 1 s a flash light was tuned on in his spacecraft after 1 second from turning on his flash light >> "t = 2s " where would be the position of the photon , will it be 300 000 km away from the spaceship and if yes dosent it mean that light is at 500 000 km away from the position it had before at "t =1s because simply the spacecraft is moving at the same time of light movement at 200 000 km/s "

I didn't look at your diagram, but what you say here is basically correct. Please see my last post for what I hope is a clearer version of the same situation.

 

[firavia]

Now I finnaly Got it thanks DOC Al thank u very much .

 

[Bob S]

Here is a real world example. At Fermilab, west of Chicago, there is a superconducting ring (proton synchrotron) about 4 miles in circumference called the Tevatron. In the Tevatron, there is a beam of protons with an energy of about 980 GeV (over 1000 x proton's rest mass), so the velocity is 0.999999541 times the speed of light. Simultaneously, in the other direction, there is a beam of antiprotons, also traveling at 0.999999541 times the speed of light. These two beams collide at specific points around the ring. What is their relative velocity? Is it greater than the speed of light?

 

[HallsofIvy]

Doc Al gave the correct formula in his first post in this thread. Didn't you read it?

$$\frac{0.999999541c + 0.999999541c}{1+ \frac{(0.999999541c)^2}{c^2}}$$
$$= \frac{1.999999082c}{1+0.999999082000210681}$$$$= \frac{1.999999082c}{1.999999082000210681}= 0.99999999999989465945164869940338c$$

 

[firavia]

I have another question about simultaneity , as we know simultaneity consist on the nonoccurence of 2 events at the same time in 2 different frame of reference , in one reference of frame 2 events may occur at the same time but in another frame of reference these 2 events are not goint to occur at the same time , if this is true I am forced to conclude tha time dilation equation is broken due to the following reasons:
a man inside a spacecraft is traveling at 200 000 km/s reached Earth at t =1s according to a terran observer time , at this moment the observer has turned on 2 flashlight at simultanely , so if we calculate the time at which the 2 flashlight were detected according to the astraunot's frame of reference we use time dilation equation so for both flashlights according to the astraunot must happen at the same time as well cause the equation for both flash light include the same constants .

but simultaneity say otherwise , as we know simultaneity requires that one of the flahs lights is seen before the other by the astraunot's frame of reference and by that the time calculated for detecting both flash lights according to the astraunot's frame reference differed though the t = t0 / radical (1-(v square/c saqure) ) is used for both flashlights to determin t0 >> the time of occurence according to the astraunot's frame of reference and t >>> the time of occurence of both flash lights according to the terran observer is the same for both flash lights ?
hoepfully U understood me " sorry for my bad english " I am a foreign student

 

[Doc Al]

I have another question about simultaneity , as we know simultaneity consist on the nonoccurence of 2 events at the same time in 2 different frame of reference , in one reference of frame 2 events may occur at the same time but in another frame of reference these 2 events are not goint to occur at the same time , if this is true I am forced to conclude tha time dilation equation is broken due to the following reasons:
a man inside a spacecraft is traveling at 200 000 km/s reached Earth at t =1s according to a terran observer time , at this moment the observer has turned on 2 flashlight at simultanely , so if we calculate the time at which the 2 flashlight were detected according to the astraunot's frame of reference we use time dilation equation so for both flashlights according to the astraunot must happen at the same time as well cause the equation for both flash light include the same constants .

but simultaneity say otherwise , as we know simultaneity requires that one of the flahs lights is seen before the other by the astraunot's frame of reference and by that the time calculated for detecting both flash lights according to the astraunot's frame reference differed though the t = t0 / radical (1-(v square/c saqure) ) is used for both flashlights to determin t0 >> the time of occurence according to the astraunot's frame of reference and t >>> the time of occurence of both flash lights according to the terran observer is the same for both flash lights ?
hoepfully U understood me " sorry for my bad english " I am a foreign student

I'm not quite clear on the set up of your question or your application of time dilation, so you may have to rephrase it. Let me say a few words about simultaneity. Whether or not the astronaut sees those two flashlights flash at the same time depends on the setup. If the flashlights are at different locations along the line of motion of the astronaut, then the astronaut will see them flash at different times. (He'll claim that the one he passes first was the last to flash.)

Time dilation and the issue of simultaneity are two different (but interrelated) "effects" of relativity, so be careful not to mix them up. (Length contraction is a third effect.)