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Don't fully get Lorentz equations

  1. Jul 16, 2011 #1
    ok - i do understand roughly what their purpose is, i think that if you have an object, a traveling at half the speed of light with constant velocity relative to another object, b which is at rest for 2 seconds, then, due to the effects of time dilation, the clock of a will slow so they actually travel for root 3 (or 1.73) seconds. However, you could consider either object to be at rest, surely, so how would you work out which object's clock would move slower (i.e. b is moving at half the speed of light in the opposite direction relative to a which could also be considered to be at rest)? Or would both objects think the clock of the other object is moving slower, since relative to them, the other object is moving at half the speed of light (in one direction or another), which would probably raise more questions...
    hmm - i think i get this part now, but still don't get the length contraction bit below so any help with that would be great

    Also, when i sub in values for the length contraction equation, i just get 0.
    On the top, i get... 1 (light second traveled) - 0.5 (light seconds / second) * 2 (seconds) which = 0. Am i subbing in a value wrong or something? Should x be the length of the object to find out its length contraction, x prime, or something? Even if that is right, if the object was 1 light second long then why would it have 0 length? Knowing me, i've probably just done something stupid, though

    Cheers
     
    Last edited: Jul 16, 2011
  2. jcsd
  3. Jul 16, 2011 #2
    The clock that accelerates (feels g-forces) will go slower than the clock that doesn’t. Also, length contraction is independent of time. The value should be the same as that for time no matter how much time has passed.
     
    Last edited: Jul 16, 2011
  4. Jul 16, 2011 #3

    bcrowell

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    This is true if the clocks are going to be synchronized, separated, and then reunited, and that's how you're going to judge time dilation. However, that isn't really what the Lorentz transformation equations are directly talking about.

    Yes, that's right. To me, that's more of a direct statement of the literal meaning of the Lorentz transformation equations.

    I'm not sure what you mean by "independent" here. Do you mean that gamma is not time-varying, because the v in the Lorentz transformation equations is taken to be a constant?

    Could you format some equations in LaTeX for us, preferably using symbols rather than numbers? I'm not having much luck understanding your reasoning from the above description. Here's an example of how to express one of the equations using LaTeX:
    [tex]x'=\gamma x-\gamma v t[/tex]
    To see how I did that, click on the QUOTE button on this post. This is in units with c=1.
     
  5. Jul 16, 2011 #4
    Yes, that's what I meant.
     
  6. Jul 16, 2011 #5

    ghwellsjr

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    Yes, you got everything right so far, you answered your own question correctly, as long as you realize that the value of gamma is 2 divided by the root of 3 which equals 1.1547. So to calculate how much time has progressed on a's clock during the 2 seconds that he travels according to b, you would divide 2 seconds by 1.1547 which equals the same number you got 1.732.
    I'm not sure what equation you are using or how you did your calculation but all you need to do is calculate the distance that b observes of a's travel which will be 2 seconds multiplied by his speed of 0.5 light-seconds/second for a distance of 1 light-second. That's from b's perspective. To get the same distance from a's perspective you divide 1 light-second by gamma which gives you 0.866 light-seconds.

    Now you can see that a will verify that his speed is equal to his distance traveled, 0.866 divided by the time it took, 1.732, which gives 0.5 c.
    There is no acceleration in this scenario so each observer will see the other one's clock as running slower and the other one's length along the direction of travel as shorter.
     
    Last edited: Jul 16, 2011
  7. Jul 16, 2011 #6
    Yes, but the question was asked as to which clock has actually slowed down. That depends upon which traveller has accelerated since one of them must have because of the velocity difference. The question as to what each traveller thinks, according to the SR equations, is surely a different one.
     
  8. Jul 16, 2011 #7

    ghwellsjr

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    No, even if we know that a and b were at rest with one another and that a accelerated, there still is no truth to the idea that one of the clocks is ticking at a slow rate in any actual sense.
     
    Last edited: Jul 16, 2011
  9. Jul 16, 2011 #8
    You're right: traveller 'a' doesn't see his own clock ticking slower during the journey. But after 'a' has decelerated so that a and b are at rest again, it will be seen that the clock on 'a' will show an earlier reading than that on clock 'b'.
     
  10. Jul 16, 2011 #9

    ghwellsjr

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    But this is only true in certain frames, like the frame in which they both start out and end up at rest. There are other frames in which this isn't true. We only consider those attributes that are true in all frames to have any sense of "actual" to them.
     
  11. Jul 16, 2011 #10
    firstly, cheers for the replies so far, guys.

    Yeah. I'll try to make what I was trying to do a little clearer now,
    [tex]x'=\gamma x-\gamma v t = \gamma (x - vt)[/tex]
    then subbing in some possible values for x, v and t:
    [tex]x'=\gamma (1 - 0.5*2) = 0 * \gamma[/tex]
    (with units c=1, of course), so you've got an object traveling at half the speed of light for 2 seconds, and the object is 1 light second long... (thinking i may have the wrong idea about what x, v or t means (probably t), since why would the amoung of time it's traveling for have anything to do with its length contraction?)

    - just incase I'm wrong about one of these, I think x is basically proper length (i.e. rest length of an object), x' is the one with the effect of length contraction, t is proper time (so like the time the object observing it measures), t' is the time with the effects of time dilation, and v is velocity, which would be a constant and gamma = 1/root[1-(v^2)/(c^2)]

    Also, as a side thing, according to:
    http://www.youtube.com/watch?v=aJVnXHKSVbw&feature=related
    http://www.youtube.com/watch?v=YAIhP2SNLDs&feature=related
    (you can see this at around times of 2:50 and 3:50 respectively they get)...
    [tex]x'=x [divided by] \gamma[/tex]
    so in their example where x = 500m and v = 0.75c, so x' = 330.7m

    and [tex]t'=t\gamma[/tex]
    so in their example where t = 80s and v = 0.8c, so t' = 133s

    - are these correct? - i thought that the lorentz equations were [tex]t' = (t - (vx/c^2))\gamma [/tex]
    and [tex]x' = \gamma(x - vt)[/tex]
    as opposed to what that was saying.

    - if they are, then an object moving at 0.5c for 2s and it's 1 light second long.
    - therefore you'd get 0.866 light seconds long as opposed to the 0 I was getting, from the 1 light second at rest and 2.309s as opposed to root 3 or 1.732s I was getting from the 2s (which seems to make sense, certainly a lot more than our answers).
    - I'm half expecting to find out they're both right and im somehow not understanding one of them properly

    - I hope that makes sense and you get where im going wrong now....
     
  12. Jul 16, 2011 #11
    OK let us assume that you are using this equation:

    [tex]x' = \frac{x-vt}{\sqrt{1-v^2/c^2}} [/tex]

    It looks like you have pugged in x=1 for distance travelled, v=05 and t=2 for the time it took to travel that distance. The mistake you have made is not realising that x and t are coordinates of events. The way to do it is this. Let us assume in the unprimed frame that the spatial coordinate of one end of the rod is x1=0 and the other end is x2=1. Now to make any sensible length measurement we should make the measurement of both ends simultaneously so that t1=t2 and in this case t1=t2=2 seconds. Now we can work out what the length of the rod is according to the observer at rest in the primed frame thus:

    [tex]x' = \frac{x2-v*t2}{\sqrt{1-v^2/c^2}} - \frac{x1-v*t1}{\sqrt{1-v^2/c^2}} = \frac{(x2-x1) -v(t2-t1)}{\sqrt{1-v^2/c^2}} - \frac{1-0.5*0}{\sqrt{1-0.5^2/1^2}} = 1.1547 [/tex]

    The result is larger than 1.0 because it is the proper length of the rod.

    If the measurements are made simultaneously, then t2-t1 is always zero so you can simply use:

    [tex]x' = \frac{x}{\sqrt{1-v^2/c^2}} [/tex]

    and if t2'=t1' you can use:

    [tex]x = \frac{x'}{\sqrt{1-v^2/c^2}} [/tex]
     
    Last edited: Jul 16, 2011
  13. Jul 17, 2011 #12
    Thank you everyone who replied, that was very helpful. I think I get it now. So... the proper length (which will have the largest value) would be measured by any observer which measures both ends of this object at the same time irrespective of their velocity.
    With proper time then, is the smallest value for time which would be measured, and would be measured by the observer which is on the object (i.e. root 3 in the example from my first post), and if the proper length was 1 light second, then the length measured by the observer moving on/with the object would be 0.866 light seconds long, (or [root (3)]/2) (assuming t'1=t'2) as stated by ghwellsjr

    - provided that those points I just said are correct, I recon I've got it.
     
  14. Jul 17, 2011 #13

    ghwellsjr

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    No, you don't get it.

    Proper time is what an observer sees and measures with his own clock at rest with him. Proper length is what he sees and measures with his ruler at rest with him. If he sees and measures a clock that is traveling with respect to him, it will be time dilated, meaning its ticks will take longer than the ticks on his own clock. If he sees and measures a ruler that is traveling with respect to him, it's length will be contracted along the direction of travel, compared to his own ruler.

    Look at what you said at the end of your post:
    "if the proper length was 1 light second, then the length measured by the observer moving on/with the object would be 0.866 light seconds long"​
    You said the proper length was 1 light second but then you said it was 0.866 light seconds.

    In your original scenario, you never said there was a length of anything, you just said a traveled for 2 seconds at 0.5 c according to b. I showed you how to calculate the time and the distance that a would then see and measure. Those are the proper time and proper distance for a.

    But then you said in post #10 that a was an object that was 1 light second long.
    "so you've got an object traveling at half the speed of light for 2 seconds, and the object is 1 light second long..."​
    That's longer than the distance a traveled. So I'm not sure we have cleared you up on all the points of concern that you have. Maybe you should restate your scenario providing all the information that is relavent.
     
    Last edited: Jul 17, 2011
  15. Jul 17, 2011 #14
    Yeah, I in post #10, the situation I stated changed slightly, but does it really matter if you're measuring a distance between 2 points in space, or the length of an object (which is also, surely the distance between 2 points). Wouldn't the length contract the same amount? If so, then it shouldn't really matter which you're using imo.

    With the proper time and length, I guess I might have misunderstood what they were. I wasn't too sure, so I just tried to look it up and found:
    http://www.mta.ca/faculty/Courses/Physics/4701/EText/Proper.html [Broken]
    which is on the website of some uni or other and it seems to have other ideas to you about what proper time and length are (although I that's not really my issue just now, I don't really care if I don't understand what proper time and length are just now).

    Where you say that I've given the proper length to be 1 and 0.866 light seconds, I meant that the stationary observer which the object is moving 0.5c away from gets the length to be 1 light second, and the one which is moving with / on the object gets it to be 0.866 light seconds. Since you defined proper length as being "Proper length is what [an observer] sees and measures with his ruler at rest with him", by an observer, I'm assuming you mean one which the object is moving away from, not one which is moving with the object and simply considering itself to be at rest. If so, I don't really see why you thought I gave 2 different values for the proper length, since the second is of an observer moving with the object, as opposed to one which is moving at c/2 relative to the object.
     
    Last edited by a moderator: May 5, 2017
  16. Jul 17, 2011 #15
    ok - I'll try to restate a couple of examples now, and explain what I currently think the solutions would be. I'll attempt to be clear... (atm I'm going with the definitions of proper time and length which are on http://www.mta.ca/faculty/Courses/Physics/4701/EText/Proper.html [Broken])

    There's my origional example for time dilation:
    You have an object moving at c/2 relative to a stationary observer (b). It has another observer (a) on it. This object is observed for 2 seconds, so moves 1 light second, as observed by b. Therefore, if I'm right about the proper time and length, the proper time will be measured by a and the proper length by b

    What I thought you should do (and seemed to be confirmed by ghwellsjr) is:
    Using the equation [tex]t' = \frac{t-vx/c^2}{\sqrt{1-v^2/c^2}} [/tex]
    you sub in values and get [tex]t' = \frac{2-0.5*1/1^2}{\sqrt{1-0.5^2/1^2}} = \sqrt{3} = 1.732[/tex]
    so observer a sees the object travel for 1.732s as opposed to the 2s observer b sees.

    Now it seems like I'm getting 2 different answers for the length which observer a would say that the object traveled during this time.
    according to ghwellsjr, you divide 1 by gamma to get 0.866 light seconds, but I'm not too sure what exactly he did (if you do time * speed, you get this answer, too)
    and according to yuiop you multiply 1 by gamma to get 1.1547 light seconds, and the reasoning behind what he's doing seems to make sense.
    But, if x' is the proper length (as it seemed like yuiop said), and t' the proper time, then the proper length would be measured by b, therefore, if you rearrange the equation, you get [tex]x = \frac{x'}{\gamma} = 0.866[/tex] (0.866 light seconds btw).
    - that's what I'd get for the answers and I hope my reasoning behind what I was doing made sense.
     
    Last edited by a moderator: May 5, 2017
  17. Jul 17, 2011 #16

    ghwellsjr

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    The article that you linked to in your previous two posts is total nonsense.

    Consider two observers each with the own clock that start out at the same place and at the same time and their clocks have the same time on them. They are both going to travel to a distant location and arrive at the same moment. One of them takes a straight line path at a moderate speed. The other one takes a round about path at a much higher speed. Both of their clocks keep proper time but when they arrive at their destination, their two clocks differ. The one that traveled at the higher speed will have an earlier time on it.

    By contrast, your article implies that there is only one time that is the correct proper time.

    Proper time is simply the time that a clock keeps track of, no matter how the clock moves. Trying to define proper time in terms of two events and reference frames is wrong.

    I don't have time right now to respond to your other points but I'll get to them later.
     
  18. Jul 17, 2011 #17
    I probably did not make it clear in my previous post that it is not always x' or more properly [itex]\Delta x'[/itex] that is the proper length. The proper length of a rod is length measured in the primed or unprimed frame where the rod is at rest. Defining the proper length between two events where no material object is involved is a bit more tricky, but I guess you could define it as the length measured in the reference frame where the spatial separation is largest.

    Similarly the proper time is the time measured in a reference frame where the clock is at rest and this can be either t or t'. For example if [itex]\Delta x' = 0[/itex] then [itex]\Delta t'[/itex] is the proper time interval and if [itex]\Delta x = 0[/itex] then [itex]\Delta t[/itex] is the proper time interval.
     
  19. Jul 18, 2011 #18
    alright. cheers. ill ignore what that article said about proper time and length then, but anyway, I don't quite get why you did distance / gamma to get 0.866, ghwell. Like, what equation were you using?
    I think I followed what yuiop did to get 1.15 light years, and it seemed to make sense, but then if you do distance / time, to work out the speed of the object, you get 4/3c, as opposed to 1/2c which is what's confusing me. (I think I worked out what t' is correctly)
     
  20. Jul 18, 2011 #19

    ghwellsjr

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    I took a short cut to get 0.866 light seconds for the distance a traveled during your scenario. This is the correct answer, by the way, and I explained how I did it in post #9.

    However, at that time, I wasn't sure what equation you were using because you called it the "length contraction equation" which yielded an answer of 0 and so I didn't understand what you had done. Now that I have had more time to go back and study this thread, I can see where your problem is.

    Going back to your first post, you correctly calculated using the Lorentz Tranform, the time that a has on his clock corresponding to the event when b's clock reads 2 seconds.

    And you also correctly calculated the x-coordinate of a's position corresponding to the same event.

    In yuiop's first post, he explained that the Lorentz Transform deals with coordinates of events. Let me expand on his brief comment:

    In Special Relativity, we define a scenario in terms of a particular Frame of Reference. You did this in your first post when you stated that b was at rest and that a traveled at 0.5c for 2 seconds. You are describing two events in b's FoR (where he is at rest). The first event is the beginning of your scenario when a and b are together and their clocks both read 0. Since we are doing everything in the x direction, we can ignore the y and z components and refer to the [t,x] coordinates for this event as [0,0] in b's FoR. The second event is the end of the scenario at 2 seconds and is [2,0] in b's FoR.

    Now what you correctly did was transform these two events into a's FoR. You got [1.73,0] but you didn't believe this because you incorrectly thought the x component was supposed to give you the distance that a traveled. Remember, a FoR is defined for an inertial object for the condition that the object is at rest, in other words, its spacial coordinates will always be zero.

    So if you want to know how far a "traveled" in a's FoR, you need to transform b's final event into a's FoR and see how far b appears to have traveled according to a. This event is [2,0] in b's FoR and [2.31,-0.866] in a's FoR. (Do the calculations and see if you get the same answers.) So a perceives b to have traveled 0.866 light seconds behind him.
     
  21. Jul 19, 2011 #20
    ooohhh... that clears up some stuff - (x is 0 for b because it will be the origin of its frame of reference, and same for a), I don't quite get your answers, but I can nearly get them...

    So. Because we're in b's frame of reference and want to find the distance between b and a, we use t=2 (the time between initial and final events) and x=0 (the position of b) as the final values (at the instant we want to find the distance between the two observers). Because a moves at 0.5c relative to b, we plug in 0.5 as the velocity (with units c=1).

    [tex]t' = \frac{t-(vx)/c^2}{\sqrt{1-v^2/c^2}} [/tex]
    sub in values stated above:
    [tex]t' = \frac{2-(0.5*0)/1^2}{\sqrt{1-0.5^2/c^2}} = 2.31[/tex] (like you got)

    [tex]x' = \frac{x-vt}{\sqrt{1-v^2/c^2}} [/tex]
    plug in values as above:
    [tex]x' = \frac{0-0.5*2}{\sqrt{1-0.5^2/c^2}} = -1.15[/tex] (not what you got, but if you do distance / time then you do get 0.5c as the speed b will have observed a traveling at...)
     
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