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  1. If you expose frozen chips to air, more moisture will condense on the chips from it. Every ml of water condensed from the air adds 1g to the chips.. The chips, after baking, lose moisture, which concentrates the calories, so if you now weigh 100g of baked chips there are more calories.
    4 points
  2. Hasn't the OP just proved (in a rather long-winded way) that the inner product of four acceleration and four velocity is zero and, hence, that if the four velocity is \( V^t=c \), \( V^x=V^y=V^z=0 \) then the four acceleration has \( A^t=0 \) (and that the four acceleration is spacelike if it isn't zero)? They just appear not to have noticed that in general \( \frac{dA^t}{d\tau} \neq 0 \), so the conclusion that \( \gamma=\mathrm{const} \) does not follow. The fact that a function is instantaneously zero does not mean that either its integral or its derivative need be zero. To me, the argument looks similar to saying that in circular motion in the x,y plane there comes a point where \( (v_x,v_y)=(v,0) \), and at that point \( a_x=0 \), and hence \( v_x \) can never change and circular motion is impossible. Obviously that's nonsense, and realising that \( \frac{da_x}{dt}\neq 0 \) is a part of understanding why it's nonsense.
    3 points
  3. Not that I know of. You do have a section to test your LaTeX beforehand, as you already know: https://www.scienceforums.net/forum/99-the-sandbox/
    1 point
  4. Seeing your 'signature', reminds of this one, which may seem apt with my original query :- Confucius also says "He who asks a question is a fool for five minutes; he who does not ask a question remains a fool forever." 🙂
    1 point
  5. (7) does not follow from (6). The term in square brackets in (6) is symmetric in \( \alpha \) and \( \beta \), so (6) is satisfied by any arbitrary four-index tensor \( R_{\alpha\beta\gamma\delta} \) (not just the Riemann tensor) as long as it is antisymmetric in \( \alpha \) and \( \beta \). This is not surprising, since (6) is nothing more than a restatement of (4), which itself is nothing more than a statement that \( R \) is antisymmetric in its first two indices (true of the Riemann, but also of a great many other tensors). It certainly does not follow that all components of the Riemann tensor (or any other arbitrary four index tensor antisymmetric in its first two indices) are zero, as can be verified by (for example) calculating the Riemann tensor for the Schwarzschild metric and confirming that (6) is satisfied for at least one choice of \( \gamma \) and \( \delta \) for which at least some of the \( R_{\alpha\beta\gamma\delta} \) are non-zero. E.g: pick \( \gamma=r, \delta=t \). The non-zero \( R_{\alpha\beta r t} \) in (+---) signature Schwarzschild coordinates are \( R_{trrt}=2m/r^3 \) and \(R_{rtrt}=-2m/r^3 \). Thus (6) expands to \[ R_{trrt}(g^{t\mu}g^{r\nu}+g^{r\mu}g^{t\nu})+R_{rtrt}(g^{r\mu}g^{t\nu}+g^{t\mu}g^{r\nu})=\frac{2m}{r^3}(g^{t\mu}g^{r\nu}+g^{r\mu}g^{t\nu}-g^{r\mu}g^{t\nu}-g^{t\mu}g^{r\nu}) \]The term in brackets on the right hand side is clearly zero, so (6) is satisfied even though the Riemann tensor is clearly not zero.
    1 point
  6. Yeah, nice insight. \( \gamma^{-1} \frac{d\gamma}{d\tau} \) doesn't have to be zero, even though \( \gamma^{-1} \frac{d\gamma}{d\tau} - \gamma^{-1} \frac{d\gamma}{d\tau} \) is identically zero. Welcome to the forums, @Kino.
    1 point
  7. Beautifully explained, thank you. And makes perfect sense when you think of burst water pipes because the ice water expands as it defrosts. And your explanation of the calorie increase is logical too, so thank you for that also String Junky Lord E
    1 point
  8. The answer for your initial question is:
    1 point
  9. Quite possibly so! To be honest, I am still struggling to understand what it is the OP is actually trying to do. It also seems he has abandoned this thread, just like all the other ones he opened before this. Indeed - very well summarised! +1
    1 point
  10. To us, yes, but I've heard professional comedians complain that it's too much too fast. You update your dead baby jokes and QAnon has moved on to pedophilia, so you stay up all night writing but nobody's listening to your act because now they're talking about lizard people taking over. The speed of crazy exceeds the speed of funny, sadly.
    1 point
  11. You may be right. I'm not following this thread very closely, and I'm not sure if what you say is what the OP is trying to prove. But here's a more standard proof: This is obvious, but let's do a check. And gammas, of course, are in general time-dependent. SR can deal with accelerations, as Markus said. The 4-vectors are, and their 4-product is, It's necessary to keep in mind that, The derivative of the gamma is, So the 4-product is indeed identically zero: As Markus also said, the concept that in SR supersedes constant acceleration is that of hyperbolic motion. He also has been very careful to distinguish flat space-time from prescription to adopt inertial frames. Indeed, the Minkowski spacetime can be studied in terms of Rindler charts --hyperbolically accelerated frames--. It is not difficult to show that when the motion is completely collinear (spacial 3-vectors of velocity, acceleration, and force). It's in wikipedia, although the proof is not complete, and I can provide a completion, if anyone's interested. As I said, I'm not completely sure that what I'm saying is relevant to the discussion. It is the standard, reliable, mainstream formalism that we know and love.
    1 point
  12. I am uncertain what it is really is that the OP is trying to show, so I can’t guarantee this. What I presented is my own understanding of what he has posted. Yes of course, but that isn’t what the OP has been doing. That is kind of my point. He’s essentially using an inertial coordinate system to show that there is no acceleration - which is trivially true. Ok, but then, what is his point? He starts with a metric in Cartesian coordinates, then manipulates it using relations that imply an inertial observer, and ends up with the conclusion that there is no acceleration...? I’m not sure I understand your question. Both charts cover the same spacetime, so the difference is merely one of coordinate basis. Rindler is what you naturally get when you apply the proper transformations that arise in a frame with non-zero constant acceleration (since such frames are not related by Lorentz transformations): \[ct\rightarrow \ \left(\frac{c^{2}}{a} +x’\right) sinh\left(\frac{at’}{c}\right) \] \[x\rightarrow \ \left(\frac{c^{2}}{a} +x’\right) cosh\left(\frac{at’}{c}\right) -\frac{c^{2}}{a} \] The advantage here is that the world line of such a uniformly accelerated particle is one of constant x (in this coordinate basis!) as t “ticks along”, so in some sense a particle in this coordinate frame is “at rest”, even though it is uniformly accelerating. It’s the most natural choice for uniformly accelerated motion. Very simply put, the line element \(ds^2=...\) does exactly what it says on the tin - it is an infinitesimal section (element) of a curve in spacetime. To obtain the entire geometric length of a curve C, you integrate: \[L=\int _{C} ds\] To describe a uniformly accelerated particle, you can do one of two things: 1. Keep the line element in Minkowski coordinates, but make C a hyperbola to describe the motion 2. Use the line element in Rindler coordinates, which makes C a “straight line” in these coordinates Of course, the result will be the same (since this integral is an invariant), so both are valid, but the computational effort differs - it turns out that option (2) is very much easier to do. It’s simply the more natural choice - in much the same way as (e.g.) spherical coordinates are the natural choice to describe the surface of a sphere, rather than Cartesian coordinates.
    1 point
  13. Just the fact that you use 'Conservative' interchangeably with 'Republican' is worrying. Conservatives in Canada and Europe are more progressive than American Democrats. At least half of American Republicans are just plain nuts. The implication is that some of us who tend to be a little fiscally conservative are interchangeable with American Republicans. You have used that wide brush on me, a few times … And used perceived, or imagined, offense to unfairly lash out at me. ( bottom of page 3 )
    1 point
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