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Time dilation limit? (how much does time dilate at c )

  1. Sep 3, 2008 #1
    time dilation limit? (how much does time dilate at "c")

    Hi, this is my first post here. I'm sort of an amateur.
    Here's my question:
    I was concerned about the idea, many times suggested, that if you were to travel -and I know that it's impossible- at the speed of light, you would see that everything in the world around would slow down and eventually stop (Of corse, you wouldn't see it, but stay with me).
    I wanted to know if this was real, so I made calculations of my own, and get to the conclusion (perhaps ridiculous) that as the effect described is a consequence of time dilation, I needed to calculate how much would time dilate at the speed of light.
    I didn't knew how to use the Minkowski diagram to do this, so I use a method of my own but I came with a way that seems right to me, and I found that there is a maximum top of time dilation and it's as short as a ratio a little shorter than 1:4. Meaning, if that's right, that time can only dilate less than four times with respect to that of the other frame of reference. In other words, every second of one frame stands for four of the other.
    This would be the maximum possible dilation.
    Please tell me if I got everything completly wrong.
     
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  3. Sep 3, 2008 #2

    Doc Al

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    Re: time dilation limit? (how much does time dilate at "c")

    The "time dilation" formula is simply:

    [tex]\Delta T = \frac{ \Delta T_0}{\sqrt{1 - v^2/c^2}}[/tex]

    Where the [itex]\Delta T_0[/itex] represents a time interval as measured by the "moving" clock or process and [itex]\Delta T[/itex] represents your "rest frame" measurement of that same time interval.

    As you can see, there's no limit to the time dilation factor as the relative speed approaches the speed of light.
     
  4. Sep 3, 2008 #3
    Re: time dilation limit? (how much does time dilate at "c")

    but if I were to apply this formula using the speed of light as the velocity wouldn't it give 0 on the bottom? I mean, wouldn't "v" = "c", so both square numbers on the lower right would cancel each other and end up as 1, therefore giving the square root of 1-1. That would lead to dividing by zero, wouldn't it?
    I'm sorry, I'm not to good with formulas.
     
  5. Sep 3, 2008 #4

    HallsofIvy

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    Re: time dilation limit? (how much does time dilate at "c")

    Well, v can't equal c so that doesn't matter- except in so far as it tells you what happens for v very close to c. By taking v close to c, we can make the denominator as close to 0 as we please. What does that tell you about T? For example, what is T0/0.000001?
     
  6. Sep 3, 2008 #5

    Doc Al

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    Re: time dilation limit? (how much does time dilate at "c")

    Right. Putting v = c leads to all sorts of problems. Luckily, relativity prohibits objects with mass from reaching light speed. But you can get as close as you like.

    Thought experiments that begin by saying "imagine that you were moving at the speed of light" are non-starters since they cannot happen even in theory.
     
  7. Sep 3, 2008 #6
    Re: time dilation limit? (how much does time dilate at "c")

    Thanks for the patience.
    -I'm dumb as a rock, but...
    light doesn't have mass and it travels at c, how can it be that one can't measure it's SR properties? I'm trying to imagine light from it's POV. I know that it doesn't have eyes (nor does a ladder and yet we say that it can enter inside a garage, don't we?)
    In any case, thanks for your answers and for your extreme patience.
    Yet, if you can and you have time, would you people mind to watch my resolution and tell me where is it wrong?
    It's a bit childish, I know, but I'm looking for visual solutions as I really want to understand what's happening inside this formulas. I tried to make it as simple as possible.
    Those weird lines are the representation of moving beams of light bouncing in two mirrors.
     

    Attached Files:

  8. Sep 3, 2008 #7

    JesseM

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    Re: time dilation limit? (how much does time dilate at "c")

    In relativity when you talk about what a given object observes in its own frame, you're not talking about what it would see visually even if it did have eyes, rather you're talking about what coordinates it would assign to various events if it was using a system of rulers and synchronized clocks at rest relative to itself to assign these coordinates. For example, if I have a ruler with a clock sitting at the 3-light-year mark that is synchronized with my own clock, and an explosion happens next to that mark when the clock there reads Jan. 1 2010, then I'd assign that event coordinates (x=3 light-years, t=Jan. 1 2010) in spite of the fact that I wouldn't see the event until Jan. 1, 2013.

    The point is that although this procedure works fine for slower-than-light observers, you can't actually accelerate rulers and clocks to the speed of light so it doesn't make sense as a way for a photon to assign coordinates to events. Even if we just consider the limit as rulers and clocks get arbitrarily close to the speed of light, the problem is that in this limit the rulers would get arbitrarily close to being shrunk down to zero length while the clocks would get arbitrarily close to being completely frozen, and zero-length rulers and frozen clocks wouldn't work for building a coordinate system.

    I can't really follow what you're trying to do in your diagrams, but in your third diagram you say "as we are talking about a beam of light, its vertical is always c"--this is incorrect. The total speed of the beam of light must be c, which means that if it's moving diagonally, its speed in both the vertical and horizontal must be less than c. For example, if the light beam is going at a 45-degree angle, then in 1 second, it will travel [tex]\sqrt{0.5}[/tex] light-seconds in the horizontal direction and [tex]\sqrt{0.5}[/tex] light-seconds in the vertical direction, so that by the pythagorean theorem, the total distance it has traveled in one second is [tex]\sqrt{(\sqrt{0.5})^2 + (\sqrt{0.5})^2}[/tex] which is just equal to the square root of (0.5)+(0.5)=1, and of course the square root of 1 is 1, so the light beam has moved a total of 1 light-second diagonally in 1 second of time.
     
    Last edited: Sep 3, 2008
  9. Sep 3, 2008 #8

    Doc Al

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    Re: time dilation limit? (how much does time dilate at "c")

    Let's start with your first attachment. What you are describing is often called a "light clock", where you have a light pulse bouncing back and forth between vertically spaced mirrors. (You can use the pulses as a timer, thus it's a kind of clock.)

    Viewed from a "stationary" frame, the light just bounces back and forth vertically. But viewed from a moving frame, the light moves diagonally. (Note that it is a key premise of relativity that light always moves at the same speed c as measured from any frame.) Let's call the time that it takes for the light to move diagonally from the bottom mirror to the top mirror t seconds according to observers in the moving frame. Then the light path would form a right triangle whose hypotenuse would have length ct. The vertical side would be length L (the vertical distance between the mirrors) and the bottom side would have length vt.

    The angle of the diagonal will depend on the speed of the moving frame. As that speed approaches c, the angle with the vertical will approach 90 degrees, just like you state. So far, so good.

    But when I look at your second attachment, I'm lost. I have no idea where that new diagonal comes from or how it relates to what you described in the first attachment.
     
  10. Sep 3, 2008 #9
    Re: time dilation limit? (how much does time dilate at "c")

    Yes. 1/0 is undefined. It can be positive or negative infinity. You can get really really close to c, and then you start to experience time extremely quickly - the lifespan of the universe passes by you in the blink of an eye. But being equal to c - there's no physical interpretation of 1/0.
     
  11. Sep 3, 2008 #10
    Re: time dilation limit? (how much does time dilate at "c")

    -JesseM, sorry, I think that I have confused you with my drawings. I do get that the speed of light is measured the same in both frames.
    -Doc Al: Sorry, I reckon that I do explained things that you already know, but it's because I may have them wrong that I put them there.
    As much as I would like to put that lines at a 90 degree angle, I can't see way. I hope that is clearer what my problem is in this second attachment.
    I've tried to make it as clearer as I could.
    Again thanks a lot for the patience, I realize how boring may be explaining something that may be to basic for you.
     

    Attached Files:

  12. Sep 3, 2008 #11

    JesseM

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    Re: time dilation limit? (how much does time dilate at "c")

    This is potentially confusing--there is no absolute meaning to "close to c" since velocity is relative, and if you're moving at close to c relative to some object like the galaxy, in your frame clocks at rest relative to the galaxy will be running much slower, not faster (but due to the Doppler effect you'll visually see them running faster if they're coming towards you, and you'll see them running even slower than the rate predicted by the time dilation equation if they're moving away from)
     
  13. Sep 3, 2008 #12

    JesseM

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    Re: time dilation limit? (how much does time dilate at "c")

    It seems like you're drawing the circle to show how far the light gets in one second, and that's correct as far as it goes, but it sort of misses the point of the light clock thought experiment--the point of the thought-experiment is to show that if it takes one second for the light to go between the mirrors in the rest frame of the mirrors, it will take longer than one second to go between them in the frame where the mirrors are moving. This is because the mirrors are still one light-second apart in the vertical direction in this frame (there's no length contraction along the axis perpendicular to the direction the mirrors are moving), but now the light has to move on a diagonal and also go a distance of vt in the horizontal direction (where v is the velocity of the mirrors in this frame, and t is the time the light takes to go from one mirror to the other in this frame), such that (1 light second)^2 + (vt light seconds)^2 = (ct light seconds)^2, which insures that the hypotenuse has length ct and therefore that the light is moving at c in this frame. If you solve for t, you'll find that it works out to t = [tex]1/\sqrt{1 - v^2/c^2}[/tex] light-seconds, which is what you'd predict from the time dilation formula.

    If you try to imagine the velocity of the mirrors as v=c, then you get the equation (1 light second)^2 + (ct light seconds)^2 = (ct light seconds)^2, which isn't going to work unless you make t infinite, so you're effectively trying to draw the "hypotenuse" of a right triangle whose vertical side is 1 light-second long and whose horizontal side is infinitely long. This hypotenuse would itself look like a horizontal line, as you can see if you consider the limit as the horizontal side gets longer and longer while the vertical side stays constant--in this limit, the hypotenuse gets closer and closer to being horizontal itself.
     
  14. Sep 3, 2008 #13
    Re: time dilation limit? (how much does time dilate at "c")

    Thanks man, I do get that in order to make the formula work the line should extend to infinity (I even put that next to the last drawing in the first page), but if you see what's next you'll see my problem. And it is this:

    If the light beam takes one second to reach the mirror, and the observer is moving at the speed of light (it's a photon), shouldn't the horizontal vector be as long as the vertical one??
    --Forget the formula for a moment: If you were making the drawing for galilleo's relativity and you'd have a ball moving at uniform speed upward at 1mph, and if the boat is also moving at 1mph, shouldn't you draw both vectors with equal length in order to be able to draw the diagonal? In fact, shouldn't you (at an inertial frame) see the ball describing a 45 degree path?
    -Now, if you apply that to this case,
    BOTH HORIZONTAL AND THE VERTICAL VECTOR REPRESENT C, SO WHY SHOULD ONE EXTEND TO INFINITY AND THE OTHER BE VERY SHORT?
     
  15. Sep 3, 2008 #14

    Doc Al

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    Re: time dilation limit? (how much does time dilate at "c")

    As long as you insist on an observer "moving at the speed of light" you will get nonsensical answers. There's no such reference frame. In relativity, a photon has no reference frame.
    Yes, for Galilean relativity! But velocities add differently in special relativity--especially if one of the velocities is the speed of light.
    Ah, but you can't apply the same reasoning to the relativistic case. For one thing, understand that in relativity the speed of a photon is always c, no matter what reference frame is doing the observing. But in your example, you are adding a vertical velocity of c (the photon with respect to the mirrors) plus a horizontal velocity of close to c (the mirrors with respect to the observer) and thinking that the resultant velocity is at 45 degrees with some magnitude greater than c ([itex]\sqrt{2}c[/itex]). That obviously cannot be correct, since the speed of the photon is always c.

    What you're missing is an understanding of how velocities add in special relativity.

    (If you review what I wrote in post #8, you'll be able to figure out the angle of the photon path for any speed v. As I stated there, as the speed v gets closer and closer to c, the angle with the vertical gets closer and closer to 90 degrees.)
     
  16. Sep 3, 2008 #15
    Re: time dilation limit? (how much does time dilate at "c")

    No. Your problem appears to be that you are interpreting the light-clock diagrams as velocity vectors. They are not velocity vectors. They are pictures of the trajectory of the photon. The diagrams do not have velocity vectors in them. There is no horizontal velocity vector, nor is there any vertical velocity vector. Since the picture is not of a velocity vector, it will not be at a 45-degree angle. It should be stated that just because one person (traveling with the clock) measures the vertical component to be c, doesn't mean that another observer measures it to be c.

    These diagrams are diagrams of the trajectory. You draw them like:
    1. Draw both paddles of the light clock with the photon at the bottom one.
    2. In time t/2, the paddles have moved a distance vt/2, so draw them in their new location, a horizontal distance vt/2 away.
    3. In this time the photon has traveled from the bottom paddle to the top paddle. Draw the (diagonal) line the photon must have taken in order to accomplish this. Depending on the value of v, this may be diagonal, or more vertical, or more horizontal.
    4. Draw the second half of the motion the same way if you like. You have now plotted the trajectory (not the velocity vector) of the photon.
    5. The speed of the photon along this trajectory is always c. The TOTAL speed is always c. Not the speed of one component or the other. The total (diagonal) speed is c. Photons travel at c in all reference frames. Hence as the clock approaches c, the photon travels in a more and more horizontal direction since more and more of the total (invariant) speed must be spent in a horizontal direction to keep up with the clock.

    So if you try to imagine a light-clock traveling at c itself, it is immediately clear that this would be impossible, because the photon must travel horizontally just to keep up with the clock, and will have no excess velocity to put into a vertical component.
     
    Last edited: Sep 3, 2008
  17. Sep 3, 2008 #16
    Re: time dilation limit? (how much does time dilate at "c")

    I may missing hundreds of things here, but I don't think that I'm missing that.
    I never said that the light reached to the corner of the square that I've drawn. In fact, you can see that I've made in both drawings a circunference that goes from the horizontal vector to the vertical in order to cut the resultant.
    This has to be made at the end, if not how could I show how much distance is yet to be traveled by the photon? (as you may convey that this extra path yet-to-be-traveled is what accounts for the time dilation).

    What I think that is the problem in my understanding, is that I still find that being both speeds equal, the horizontal vector should be at 90 degrees giving a final result direction of 45 degrees. I'll try to get better this equations that you gave me, but I fail to visualize how to apply them to this diagram.
    -The fact of me sticking so much to this diagram is because it may show a real path, and angle that you should be able to find in reality, that's way I'm so stubborn, sorry for that.

    PS: I thank you a lot for all this effort to both. I don't want you to believe that I'm one of this clowns that thinks that after reading one book has discover a flaw in Einstein's thought and is heading for the Nobel. I do respect what you say, and 'm sure that it's true. It's just that I'm trying to get results in a way that I can really understand. I'm just trying to avoid repeating stuff that I don't get.
    I'll look closer to the information that you've given me.]
    cheers
     
  18. Sep 3, 2008 #17
    Re: time dilation limit? (how much does time dilate at "c")

    Both components aren't equal. They must (vector) add to c, but they are not c individually.
     
  19. Sep 3, 2008 #18

    Doc Al

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    Re: time dilation limit? (how much does time dilate at "c")

    While there are two (almost) equal speeds in this problem, they are not the horizontal and vertical components of a single velocity.

    The two velocities are:
    (1) The velocity of the photon as measured in the mirror frame: c in the vertical direction.
    (2) The velocity of the mirror frame with respect to the "rest" frame: v, very close to c in the horizontal direction.

    To find the velocity of the photon with respect to the "rest" frame, you can add those two vectors, but you must do it according to the rules of relativistic velocity addition (which differ from the rules of galilean velocity addition). When you add them properly, you'll find the photon velocity will have magnitude c (of course) and will have an angle that depends on the exact velocity of the mirror frame.

    But you certainly don't need to know the details of relativistic velocity addition to understand the "light clock" example and the diagonal path of the photon. It's much easier than you think. Read this: Special Relativity and the Light Clock
     
  20. Sep 3, 2008 #19

    JesseM

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    Re: time dilation limit? (how much does time dilate at "c")

    If the mirrors are moving at the speed of light in the observer's frame, then why do you say "if the light beam takes one second to reach the mirror"? As I said, if the mirrors are imagined to be moving at the speed of light then the time for the light beam to go between them in this frame would necessarily be infinite.
    Sure, but if the ball is moving upward at 1 mph in the boat's frame, and the boat is moving horizontally at 1 mph in our frame, then in our frame the ball will be moving faster than 1 mph (it will be moving at [tex]\sqrt{2}[/tex] mph). Also, in our frame the ball will take exactly the same amount of time to travel between the two mirrors (or whatever it's moving between) as it does in the boat's frame. Those are the two critical differences between the ball in Newtonian physics and light in relativity--in the case of light the speed must be the same in the observer's frame as in the mirrors' frame, and therefore the time must necessarily be greater in the observer's frame than in the mirrors' frame.
    Because of the two basic differences between relativity and Newtonian mechanics that I mention above. Again since the time for the light to go between the two mirrors is no longer 1 second in relativity, if we represent the time as t, then the horizontal leg must have a length of vt light-seconds (because the mirrors have been moving at speed v for a time of t) while the vertical leg must still have a length of 1 light-second, and in order for the photon to still be moving at t we must assume the hypotenuse has a length of ct. As v gets closer and closer to c, the only way to make this work out is to make t larger and larger, so the horizontal leg keeps getting longer as the vertical leg stays a constant length.
     
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