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I Do space & time trade off for each other?

  1. May 11, 2016 #1
    Two questions for you lovely and helpful forum members:

    1. I have two clocks far from a massive object, relative to which I am at rest, and I let go of one of the clocks. I let it free-fall for 186,000 miles, and then I observe both clocks. Is the falling clock observed to be one second behind my local clock?

    2. Slightly related: As I understand it, the sum of gravitational and potential energy of an orbiting body is invariant. In the Sun's rest frame for example, this sum could be measured as a certain number of joules; however in a different frame, free-falling alongside the orbiting body, the sum would be much smaller or zero, but similarly invariant regardless. Is this considered an example of gauge invariance?

    Thank you!
     
    Last edited: May 11, 2016
  2. jcsd
  3. May 11, 2016 #2

    PeterDonis

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    First, what do you mean by "observed"? I assume, from the fact that one second appears in your question, that by "observed" you mean that you see light coming from the falling clock, carrying an image of its reading, and you compare that image with the reading of your local clock at the same instant that you see the image.

    Second, how are you measuring the distance the free-falling clock travels? Evidently you don't intend to use the reading on the falling clock in the image you see, compared with the reading on your local clock, as a measure of that distance, since your question treats the distance as independent of the observed clock readings. The simplest assumption is that there is another observer, also at rest relative to the massive object, who is below you, and you have extended a very long ruler between you and him, with the ruler at rest relative to both of you, and have verified that the distance between you according to that ruler is 186,000 miles. The image of the falling clock that you would compare with your local clock's reading is then the image of that clock that is emitted just as the clock passes the other observer.

    Given those assumptions, the answer to your question is, in general, no, the difference in clock readings will not be exactly one second; it will be longer. How much longer depends on how close you are to the massive object and how massive it is; the closer and the more massive, the longer it will be.

    What is "gravitational" energy? Usually that term means the same thing as potential energy. I think you mean the sum of kinetic and potential energy is a constant of the motion; this is true for an object in a free-fall orbit, yes, but only in a particular set of coordinates. See below.

    First, that's not what "invariant" means. "Invariant" means the quantity does not change when you change coordinates. The sum of kinetic and potential energy is, as I said above, a constant of the motion--in other words, it doesn't change as the body moves. But if you define this constant as "the sum of kinetic and potential energy", then it is not invariant, because it changes when you change coordinates.

    Actually, an even stronger statement is true: except in a particular set of coordinates, the "potential energy" of the object isn't well-defined at all. The kinetic energy is always well-defined, and changes as you change coordinates, as you note. But the potential energy is only well-defined in coordinates in which the central body (the Sun, for example) is at rest. (Strictly speaking, the central body also has to be the only body with significant gravity in the system, which is not true of the Sun. But we can pass over that complication for now.) So there is no way to define a "sum of kinetic and potential energy" that is invariant under general coordinate transformations. (But see below.)

    No. Invariance under general coordinate transformations can be considered a form of gauge invariance, but that kind of invariance is not what you were describing. See above.

    It is true that there is an invariant that can be defined for a body in a free-fall orbit about a central mass (strictly speaking, only in the idealized case of a single central mass with no other gravitating bodies in the system, as mentioned above). But this invariant is not defined as "the sum of kinetic and potential energy"; it just happens to equal that sum in coordinates in which the central body is at rest. The actual invariant is usually called "energy at infinity", and its definition is somewhat technical, but I'll give it: it is the inner product ##P_a T^a## of the object's 4-momentum ##P_a## with the timelike Killing vector field ##T^a## of the spacetime. (The latter vector field is only present in systems with time translation symmetry, which is where the restriction to systems with a single central mass comes from.) If you're not familiar with the concepts I just used, it's worth taking some time to learn about them; I would recommend Sean Carroll's online lecture notes on GR as a reference:

    https://arxiv.org/abs/gr-qc/9712019
     
  4. May 11, 2016 #3
    Ugh -- I apologize for being so loose with my first question, and making a mistake in the second. Also these were probably beginner questions; sorry.

    First, yes I did mean kinetic energy (not gravitational energy), and having correctly assumed my error, you thoroughly cleared up my misunderstanding; thank you.

    Regarding the clock, similarly you figured out what I meant to be asking. Let's take the massive body out of it. We're in a void, and the clock we release instead has a little jet that accelerates it steadily. When it passes another observer (who can transmit the clock reading) at 186,000 miles, is the accelerated clock one second behind the inertial clock? I'm basically asking whether the speed of light serves as a "conversion factor" between space and time under these conditions, which is something I may or may not have read somewhere. Thank you again!
     
  5. May 11, 2016 #4

    PeterDonis

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    Again, I'm assuming that the difference in clock readings you are referring to is the difference between the reading of the accelerated clock seen in its image (which is emitted when that clock passes the other observer 186,000 miles away, as measured with a very long ruler between you and him) and the reading on your clock at the instant you see the image. And again, the answer is no, the difference in clock readings will not be one second; it will be longer. How much longer depends on the acceleration of the jet clock: the higher the acceleration, the longer the difference.

    Not in the sense you mean, no. The speed of light is a conversion factor between units of space and units of time. But that doesn't imply that the ratio between a given measured distance and a given measured time will be the speed of light.

    You should be able to see the difference by restating your problems using the same units for both distance and time. For example, we're in a void, and we release a clock that accelerates steadily. Another observer is a distance 1 away, and the clock transmits its reading to us when it passes the observer. When we receive the clock's transmission, will the reading we receive be exactly 1 less than the reading on our clock? It should be evident that the answer to this question has nothing to do with the speed of light; it has to do with the specific details of how the other object is moving relative to us (and with the geometry of spacetime, i.e., the effects of gravity, in general--but those are absent in the particular case just described).
     
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