studiot

It has long been known by electronics engineers that audio amplifiers and other heat producing circuitry that have undersized heatsinks and situated within enclosed cases fail prematurely through overheat stress. We do not know the mechanism(s) by which the heat is transferred from the source to the outer walls or the available rate of removal of heat through those outer walls..

There must be more implied in the question to make (allow) us to assume Gauss's Law applies to the heat flux from the emitter through to the outside of the container.

When I say time does not dilate I mean time does not dilate. The coordinates of a single event point are {x,y,z,t} in one observational coordinate frame and {x',y',z',t'} in another. The values of x',y' z' and t' in terms of x, y z and t are given in a linear transformation. Concentrating on time and considering that the t' frame is moving with velocity v parallel to the x (of the t frame) axis so we can drop the y and z T' = At + Bx; where A and B are constants So t' may be greater than, less than t or even negative depending on these constants. Constant B allows for the fact that observers in t and t' may start counting time from different zero event points. Introducing the relative velocity of the t and t' frames and doing some algebra leads us to [math]A = \frac{1}{{\sqrt {\left( {1 - \frac{{{v^2}}}{{{c^2}}}} \right)} }} = \gamma [/math] and [math]B = - \frac{{\gamma v}}{{{c^2}}}[/math] So we have exchanged constants A and B for two others, gamma and c. Gamma is the Lorenz factor and is only valid for the combination of the two frames in question c is the speed of light and is valid for all frames. This is useful as these constants apply to the transformation of all four coordinates, not only time. None of these linear transformations lead to a 'dilation' of the coordinates. Noting that t and t' refer to the same event point we can answer the question what does dilate then? I said you need two (event) points for this. If we consider the difference between two points that is (x2 - x1) and (t2 - t1) in the first frame and (x'2 - x'1) and (t'2 - t'1) in the second frame We have a length and a time difference. Many physical quantities come in two flavours like this. Electric potential and potential difference both measured in volts Temperature and temperature difference both measured in degrees Each of the two flavours have the same units but somewhat different characteristics and are used for different purposes in physics. So back to relativity, these differences are taken between the same two event points (before we had one, now we have two points) but viewed in the different frames. So we are talking about the same length and the the same time difference in both cases. If you perform the Lorentz transformations on coordinates from one frame to yield the coordinates for both points in the other and take the length and time differences in each frame you will get different numbers. So observers in each frame will evaluate these differences as being different numbers. Alternatively if we substitute in the Lorenz transformations into my difference formula above (x2 - x1) and (t2 - t1) in the first frame and (x'2 - x'1) and (t'2 - t'1) in the second frame we will obtain formulae for what happens when we consider the same time difference or distance difference (length) from the standpoint of difference frames. In other words we will obtain the formulae called time dilation and length contraction. One final note These formulae are developed using linear or 'first order' analysis. This is OK when the event points are close to each other and in particular the differences can become infinitesimals. So [math]{\rm{\Delta t}}\;{\rm{and}}\;{\rm{\Delta x}}[/math] become [math]\delta {\rm{t}}\;{\rm{and}}\;\delta {\rm{x}}[/math] This allows us to define, or assign meaning to, calculus operations involving relativity formulae.

If this is a homework problem you should start by copying it out properly as written. That means accurately and completely and including any diagrams provided. I cannot believe even the most disreputable educational establishment would have written such a question. 1m x 0.5m x 1m is not cubical. There is nowhere near sufficient information to even begin to guess at the problem.

You have asked about the popular phrase 'time dilation'. This makes it very difficult to answer because strictly speaking time does not dilate. It is the time interval or time difference or time lapse that changes. This time difference is measured between two points, just as we measure length between two points.

Are you posting this in the right thread? Local surveys of the lake should surely be discussed in the lake Balaton thread? However I did answer your questions in post#19 here in this thread. Was there something there you did not follow? Remember that my objective was to do a 'back of an envelope' estimate of how much the water surface might vary so design a suitable survey method. This estimate suggests that gravitational effects will affect the surface by less than 20mm.

Lennard Jones forces were introduced in 1924. The curve makes interactions look pretty rigid at ultra short distances.

Yes, thanks for the reply I'm aware of most of that, though I am not so slick with aggregated objects. When you want a satisfactory answer in the real world you have to work out all the bit individually and I think the aggregation can obscure some of the properties and consequences. Two points There are issues with defining an 'event' as a point in spacetime due to its pointlike nature. Under this view you are forced to separate occurrences at the point (event) from the point (event) itself. This implies that spacetime is more than a mathematical model since there is something that can be separated. This view is, however, consistent with the use of the word 'Field' here. For a field here is an indexing system, with as many dimensions as there are indices, that identifies unique objects placed at any given index set. Of necessity, the object introduce their own dimensions, additional to the index ones.

Mordred when you were inadvertently linking fields to manifolds were you actually thinking of the Lorenz group? I have always contended that as soon as you introduce the word field, in the sense used here, four dimensions is not enough. https://en.wikipedia.org/wiki/Lorentz_group

They are all pulling your leg, mate. Oops sorry just realised your leg doesn't exist so you can't feel it.

No

Regarding your second question (underlined in the quote), why do you fail to discuss this further? Regarding your first qustion about decimal representation Hypothesis : Each decimal representation of the real numbers between 0 and 1 can be arranged to correspond to a unique whole number. (This is what is meant by countable or enumerable.) For the decimal corresponding to any whole number, n, let the nth digit be an. If an = 0,1,2,3,4,5,6 or 7 then take bn = (an + 1) If an = 8 or 9 then take bn = 0 Then the decimal 0.b1b2b3.........bn....... differs from every decimal in the enumerated set in at least one place, contrary to hypothesis. Hence the hypothesis fails and the set of real decimals cannot be placed in an enumerable set.

The proper length of a lightpath is infinite since it is accomplished in zero proper time.

You are right, this year is a good year to start thinking about your longer term future, here is some standard advice. https://www.petersons.com/college-search/planning-list-tenth-parents.aspx Both previous respondents have given good advice - essentially don't narrow your options just yet. In fact what makes you think you are attracted to Science?

Maybe but what is the 'proper length' of a lightwave according to Einstein's relativity?

So what equation do you get if you substitute in

Do you know what a polynomial of degree 3 is? Can you write down the general polynomial of degree 3 (here in this thread?)

Is it difficult because the English is difficult or is it difficult because the mathematics is difficult?

How about a = 1 b = -1 Edit I think this is wrong (1)-1 is not equal to (-1)1 Sorry.

I didn't ask what you hadn't done, I asked what you had done. We don't do people's homework here, we just help them to help themselves. Are you ignoring my hint?

So how far have you got? Have you written four simultaneous equations and solved them for the constants?

Well that question you posed then was a couple of years before my time here but thank you for bringing it to my attention. If you are interested, there is a good presentation and development of this view in the book by Wilson that I posted an extract from in post#34 here and that Tim likes so much (actually I do too) even though that extract ends by declaring the Aether "superfluous" The author, BTW, was Professor of Physics at Rice in Houston, Texas. Another quote that might tickle your fancy comes from Eddington (Sir A) Edit: The explanation you received from DH in your old thread about the lightsphere is essentially the geometric method I offered in post#46 and rejected by Tim. The earliest reference to this construction was I can find was by Russell.

Pity your observation contradicts the associated quote. I'm not sure whether Maxwell discussed the Aether or not, I'm sure he would have thought about it. He did, however, produce a hexagon cell structure proposal for space to discuss the propagation of EM radiation through the vacuum. I posted a reference to the source of that a couple of years ago here. I will try to find it.

So why didn't you say so in the first place? More importantly you have omitted the most important imformation, which you clearly know. What is the fluid? what are the throughput rates? What sort of pump are you employing? What is the delivery length? This appears not to be a causal shoot the breeze question to while away the hours in a coffee house. This is an attempt to gain commercial information for free, thereby putting someone who has put in years of (study) effort out of a job. Employ a real Engineer for a proper fee. He will ask those questions above and more, and then likely start by studying a Colebrook-White chart or Moody diagram for your installation. Since you did not understand the simplest equation, I would not recommend trying the engineering for yourself.

Whilst I'm pleased that my hint helped, I glad that you worked it out for yourself. That's by far the best way and hopefully you will understand it better now so that it will 'stick'. +1