Algebraic Derivation of the Schwarzschild Dilation Function Andrew Theyken Bench1

Department of and Astronomy, Franklin and Marshall College, 501 Harrisburg Avenue, Lancaster, Pennsylvania 17604

Using only a thought experiment and Einstein’s correspondence principal, a model is derived that correctly predicts the Schwarzschild time dilation expression in limiting cases. The method requires almost no prerequisite knowledge from the student and is carried out with only algebraic techniques, allowing the introduction of a mathematical example of to introductory physics students.

01.30.la, 01.30.lb, 04.00.00

1 [email protected] Introduction Being good theoreticians, we decide to work out how long There has recently been a push in the literature1 and it should take the beam of light to reach the ceiling with no in textbooks2 to integrate the concepts of general acceleration: relativity into the physics curriculum of ∆y = cr∆t (1)7 undergraduate physics students. Such a push is ∆y certainly understandable considering the current state ∆ti = (1a) of of which “general relativity cr (GR) has become an integral and indispensable ∆t → time for light to travel a distance in inertial frame part.”3 It is somewhat, unfortunate, therefore, that i introductory physics students, both at the We are ready to get underway to conduct the experiment undergraduate and high school level are rarely, if when we get a message from the captain of the ship. He ever, presented with a mathematically motivated sends his apologies, but says we just received a distress discussion of general relativity. signal and have to begin accelerating at a constant

m acceleration of 4.5×1015 (a ) (in the +y direction). Opposition to such an attempt is certainly s 2 appropriate. Attempting to make any progress, in a Undeterred, we decide simply to redo the calculation for mathematical sense, against GR, at a level (1a) in an accelerating reference frame. appropriate to introductory students is difficult, but not impossible, if we motivate the discussion by attempting to “inform” the reader, rather than Thus, we need to find an equation that will relate how describe physical fact. In the derivation that follows, much time it will take the light to reach the ceiling, in the we will attempt to extend a general conceptual model presence of a constant acceleration. This is a little bit of General Relativity into the realm of mathematical 4 tricky to conceive conceptually, but let’s change the model formalism. from a spaceship to standing on the in the presence of a gravitational force. Also, instead of shooting light upwards, let’s throw a ball upwards and ask, in the It is important to stress that what follows is an presence of gravity, how long does it take the ball to ‘intuitive’ approximation of general relativity; yet, go ∆y ? We can make this ‘conceptual’ change because of what the model lacks in precision; it gains back in approachability. Prior to attempting this “general the Einstein equivalence principal. Now we can conceive relativistic” problem, it would benefit the student very easily that the form of the equation we need is just one immensely to see the derivation of time dilation in of the standard kinematics equations: 5 . The derivation, herein given, is 1 2 ∆y = vr∆t + ar∆t (2) implicitly based on the model of special relativistic a 2 a time dilation. Also, it would be beneficial to give a conceptual explanation of the correspondence Thus, we solve equation (2) for ∆ta using the quadratic principle6 as the reference frame of this derivation is equation: the ubiquitous accelerating gedanke spaceship. − vr + vr 2 + 2ar∆y ∆t = (3) a ar Method We chose this particular solution of the quadratic because it Imagine that we are floating in an interstellar refers to the “first” time the ball reaches ∆y on the upward spaceship, far away from any source of gravity. Also, we have just spoken to the captain, and he part of its journey. Next, we transition back to the assures us that the ship is in an inertial frame, that is spaceship, where the ball we are “throwing” corresponds to to say, all the force sensors onboard indicate there is light, so v → c . no acceleration of any kind influencing the ship. We − cr + cr 2 + 2ar∆y ∆t = (3a) are very pleased by this because we want to conduct a r an experiment to determine how much time it takes a a 8 beam of light to go from the floor of the ship to the Next, we decide to plug in numerical values for both ceiling, a height ∆y . equation (1a) and equation (3a) and find out this surprising result: −9 ∆ti → 3.33×10 s −9 ∆ta → 3.42×10 s The time interval for the accelerated frame is longer. We notice that equation (1a) and equation (2) and related to Surely, this is a mistake? We can conceive that it one another by the quantity ∆y . Thus, we substitute makes sense in the case of the spaceship, because the “ceiling” of the spaceship moves upward, thus equation (2) into equation (1a) and solve for ∆ti : making the light travel a longer distance. But, if the 1 ⎛ 1 2 ⎞ t cr t ar t (4) Einstein correspondence principle is correct, then this ∆ i = r ⎜ ∆ a + ∆ a ⎟ suggests that in the presence of gravity, time is also c ⎝ 2 ⎠ dilated! Incredulous, we go back to the computer and ⎛ ar∆t ⎞ graph equation (1) and equation (2) with the relevant a (5) ∆ti = ∆ta ⎜1+ r ⎟ numerical constants. ⎝ 2c ⎠ For reasons we will make clear momentarily, equation (5) ∆y 10 is only accurate when ∆ta is very small, in fact, when:

8 2∆y ∆ta ≤ (6) 6 c Equation (6) substituted into equation (5) 4 ⎛ ar∆y ⎞ 2 ∆ti ≤ ∆ta ⎜1+ ⎟ (7) ⎝ cr 2 ⎠ Time Equation (7) represents the gravitational time dilation of 2×10 -8 4×10 -8 6×10 -8 8×10 -8 1×10 -71.2 ×10 1.4-7 ×10 -7 our simple model. ∆y vs t. The horizontal time represents ∆y =1, the linear line represents the inertial model and the quadratic curve represents the accelerated Comparison to Exact Solution model. We might ask ourselves how our model compares to actual theoretical results from GR. To do so we introduce a It seems like at every point, save zero, the linear solution to the Einstein equations for a spherically curve for the inertial model is greater than the symmetric mass, such as the Earth, known as the quadratic curve for the accelerated model. To be Schwarzschild solution.9 sure, we squeeze the time ordinal down very small: The exact solution to the time dilation problem in ∆y 2 Schwarzschild is given by the expression: −1 2 1.75 ⎛ 2GM ⎞ ∆ti = ∆ta ⎜1− ⎟ (8) 1.5 ⎝ rcr 2 ⎠ 1.25 2GM 1 To obtain a non-imaginary result, ≤ 1; therefore, r 2 0.75 rc 0.5 we can perform a binomial expansion of equation (8) in 0.25 2GM terms of 2 . In almost all cases, we need only confine Time r -9 -9 -9 -9 rc 1×10 2×10 3×10 4×10 ourselves to the first of the expansion: ∆y vs t. The horizontal time represents ∆y =1 . ⎛ GM ⎞ The bottom curve is the accelerated model. ∆t ≈ ∆t 1+ (9) i a ⎜ r 2 ⎟ ⎝ rc ⎠ Incredibly, in the accelerated frame, or by the Furthermore, we can rewrite equation (9) in terms of the correspondence principal, the gravitationally affected variables that we were using, corresponding to acceleration frame, the time interval is longer. Time runs slower caused by gravity. To do so is a good exercise for the in the presence of gravity! Excited, we tell the student. Also, we will make the change of variable, captain and he asks us to try to relate the one time r → y . We can do this because y is simply one specific interval to the other. direction of a radial path. ay ⎛ ⎞ ∆ti ≈ ∆ta ⎜1+ ⎟ (9a) ⎝ cr 2 ⎠ We compare equation (9a) to the model we derived, cr 2 equation (7): ar ≤ (8a) ⎛ a∆y ⎞ 2∆y ∆t ≤ ∆t 1+ (7) r r i a ⎜ 2 ⎟ a∆t ≤ c (9) ⎝ cr ⎠ a The correspondence between equation (7) and Equation (8a) is substituted into equation (9) equation (9a) is really quite amazing. True, we had 2∆y ∆t ≤ (6) to use some mathematical ‘trickery’ to get the a cr equations to look like one another, but the Thus, we now understand the inequality in equation (7). mathematical form of the equations is precisely the r same (save a nonessential delta and an inequality Under limiting conditions, very short ∆ta , very small a ’s, sign). This suggests, at the very least, that the or very small ∆y ’s, equation (7) ≈ equation (9a). If any of assumptions that we made in our derivation were not those factors increase, we move away from our assumed at all baseless, and that GR really does behave in a constant acceleration and the model begins to diverge from way consistent with our assumptions. Below, we will the Schwarzschild model. think carefully about the assumptions and approximations that we made. Discussion Using just the precept of a thought experiment and the Error and Justification notion of Einsteinian correspondence, we have determined 2∆y a simple, but effective, model to explain gravitational time At equation (6) we noted that ∆t ≤ . Why? a cr dilation. Einstein would have been proud! Not only do the The point is actually very important and very results make conceptual sense, but in the limit, they revealing. We used as our fundamental model for an become the solution of the exact Schwarzschild . accelerated frame equation (2). However, there was a What is more, we developed all of the concepts with very subtle Newtonian assumption in equation (2) that we simple algebra, at a mathematical level appropriate to have not discussed until now—the concept of introductory college students and advanced high school constant acceleration. Imagine that we accelerated a students. If the instructor wished to extend the results herein presented she could certainly consider the situation m 11 rocket constantly at 4.5×1015 . If acceleration in terms of the relativistic acceleration function. Such a s 2 derivation would certainly demand much more 12 were constant, at t = 6.66×10−8 s, we would mathematical sophistication from the student, but would require far fewer assumptions, and may, in that respect, achieve the ! A nanosecond thereafter actually be clearer to a mathematically prepared student. we would surpass the speed of light. This is not allowed. Relativistically, there is no such thing as constant acceleration. Yet, this is what we used as our model. Thus, to prevent unwanted relativistic 1 James B. Hartle, Am. J. Phys. 74, 14-21 (2006). 2 side effects of non-constant acceleration, we must James B Hartle, Gravity: An Introduction Einstein's general relativity (Addison-Wesley, San Francisco, 2003) only concern ourselves when the speed of our rocket 3 is well below the speed of light, thus, very short Sean M. Carroll, Spacetime and Geometry: An Introduction to General Relativity (Addison-Wesley, San Francisco, 2004), p.vii. . Determining how short “very short” is takes 4 Paul Hewitt, Phys. Teach. 43, 202 (2005). some foresight, but can be determined as follows. 5 Raymond A. Serway and John W. Jewett, Jr., Principles of Physics: A Calculus-Based Text, 3rd edition (Brooks/Cole, USA, Recall that equation (3a) contains the 2002), Vol. 1, p.285. 6 r 2 r 10 r S. P. Drake, Am. J. Phys. 74, 22-25 (2006). quantity c − 2a∆y . If a becomes incredibly 7 We c here as a vector for clarity to the student. large, then we get an imaginary result, which, 8 We have been rather sloppy about the vector nature of physically, corresponds to the fact that the light beam acceleration, in that, as we switched between the spaceship case never hits the ceiling, because the spaceship is now and the gravitational case, we have not defined the sign of traveling faster than a light beam. We know this acceleration. It is left to the student to think out the situations carefully and assign the sign of acceleration accordingly. cannot be true because the space ship CANNOT go 9 Sean M Carroll, Spacetime and Geometry: An Introduction to faster than the speed of light equation (9); therefore: General Relativity (Addison-Wesley, San Francisco, 2004), p.217. 10 cr 2 − 2ar∆y ≥ 0 (8) For clarity, we have picked the negative sign out of the acceleration vector

11 Charles W Wheeler, Kip S. Thorne, and , Gravitation (W. H. Freeman, San Francisco, 1973), Vol. 1, Chap. 6. 12 J. E. Romain, Rev. Mod. Phys. 35, 376-389 (1963).