Watch Meteor Strike on PBS. See more from NOVA.
NOVA episode aired 27 March 2013 on PBS about the meteor observed over Chelyabinsk, Russia on Feb 15, 2013.
Thursday, March 28, 2013
NOVA: Meteor Strike
NOVA episode aired 27 March 2013 on PBS about the meteor observed over Chelyabinsk, Russia on Feb 15, 2013.
Monday, March 25, 2013
Comet 2013 A1 Close Approach to Mars
The comet 2013 A1 will make a close approach to Mars next year on Oct. 19, 2014. You can check the details of the close approach for a location on Mars using JPL's HORIZONS Web-Interface with the following settings as inputs.
The minimum distance from the given location is 0.0007764 AU at 18:43 UT. To display the data in a polar plot one needs to Altitude and Declination into the corresponding Polar Angle. For the local horizon plot the Polar Angle is the angle from zenith.
The minimum distance from the given location is 0.0007764 AU at 18:43 UT. To display the data in a polar plot one needs to Altitude and Declination into the corresponding Polar Angle. For the local horizon plot the Polar Angle is the angle from zenith.
Monday, March 18, 2013
Mar 17th LPSC Mars Curiosity Briefing
lpsc2013 on livestream.com. Broadcast Live Free
Some of the MSL investigators gave a news briefing at Lunar and Planetary Science Conference 2013 this Sunday. The main topics were the use of Curiosity's mastcam to select targets for further investigation. The camera filters allowed the creation of a 12 point spectrum of the scene imaged and the location of hydrated points within an image. The water present is located in the veins in the mudrock and appears to be hydrated calcium sulfate. It is thought to be a remnant of a wetter period in Mars' past. The DAN neutron spectrometer also showed the presence of subsurface water.
Wednesday, March 13, 2013
March 12th NASA Press Conference: Mars Drilled Rock Analysis
src: Curiosity Cam
press release MSL image archive
Yesterday's NASA press conference discussed some of the results from the analysis of the Gale Crater drilled rock powder. The x-ray diffraction analysis of the rock interior indicated the presence of clay minerals called smectites which are formed by the action of water in a nearly neutral pH environment. The SAM quadrapole mass spectrometer detected water, carbon dioxide, sulphur dioxide, oxygen and hydrogen sulphide. The deuterium-hydrogen ratio was the lowest detected so far. These results indicate the presence of an aqueous habitable environment in Mars' past. The focus of study now shifts from the search for water to a search for a habitable environment. The term habitable was qualified to mean an "environment a microbe could have lived in." Even though an environment may lack organics it may be habitable for microbial life if there are organic minerals that it can feed on and a source of carbon that it can use to form organic molecules. The presence of a "carbon signal" is dependent on three components. There has to be a mechanism for concentrating carbon. The environment has to be such that it will preserve organic material. On Earth heat can break down organic material over long periods of time. Finally, the radiation environment has to be favorable.
Monday, March 11, 2013
Finding the Roots of a 4th Degree Polynomial
Today I found a method for finding the roots of a 4th degree polynomial, x4+ax3+bx2+cx+d=0. As with the cubic equation letting x=y-y0 allows us to reduce the number of terms present and get another polynomial of the form y4+By2+Cy+D=0 that can be solved instead. If y is complex then y=p+iq and substitution into the reduced equation results in a 6th degree polynomial for p that is cubic in p2 and its roots can be found using the method for finding the roots of a cubic equation.
Given a set of four roots one can quite easily find the coefficients of the 4th degree polynomial.
The coefficients for the reduced polynomial can then be calculated.
This new set of coefficients allows us to find the coefficients of the cubic equation for p. In this case it has 6 zeros.
With p known we can find the corresponding positive and negative values of q and therefore the y values. Subtracting y0 from each y value gives the corresponding x values for the roots of the original 4th degree polynomial. There is some redundency in the solution but one gets all the original roots even if they are all rational. The reason for this is that q can be imaginary.
Given a set of four roots one can quite easily find the coefficients of the 4th degree polynomial.
The coefficients for the reduced polynomial can then be calculated.
This new set of coefficients allows us to find the coefficients of the cubic equation for p. In this case it has 6 zeros.
With p known we can find the corresponding positive and negative values of q and therefore the y values. Subtracting y0 from each y value gives the corresponding x values for the roots of the original 4th degree polynomial. There is some redundency in the solution but one gets all the original roots even if they are all rational. The reason for this is that q can be imaginary.
Saturday, March 9, 2013
Friday, March 8, 2013
Why the Sum of the Roots of the Reduced Polynomial is Zero
Zeroing the coefficient of the squared term (setting A=0) resulted in the sum of the roots being zero. There is a simple explanation for this. If we let pk=x-xk with k=0,1,2 then multiplying three monomials containing the roots together involves selecting either the unknown x or the root which is designated by 0 or 1 respectively in the table below. There are eight combinations.
There is one combination with three 0s for the x3 term, three combinations with two 0s and a 1 for the x2 term, three combinations with one 0 and two 1s for the x term and one combination with three 1s for a constant term. So when we multiply the three monomials together we get coefficients that involve the roots.
Setting the coeffient of the x2 term, A, equal to zero in the reduced polynomial also requires that the sum of the roots is also zero. If we did the same for a higher degree polynomial we would find that the sum of all the roots is also zero. Smith says that the cubic was first solved about 1515 nearly 500 years ago.
There is one combination with three 0s for the x3 term, three combinations with two 0s and a 1 for the x2 term, three combinations with one 0 and two 1s for the x term and one combination with three 1s for a constant term. So when we multiply the three monomials together we get coefficients that involve the roots.
Setting the coeffient of the x2 term, A, equal to zero in the reduced polynomial also requires that the sum of the roots is also zero. If we did the same for a higher degree polynomial we would find that the sum of all the roots is also zero. Smith says that the cubic was first solved about 1515 nearly 500 years ago.
Saturday, March 2, 2013
Cubic Equations And Their Solution
A method for the solution of cubic equations was first published by Cardano in his Ars Magna in the middle of the 16th Century but was found earlier. One can reduce any cubic equation of the form x3+ax2+bx+c=0 to a simplier form, x3+Bx+C=0, by subsitiuting x=y-y0 and choosing y0 so that the coefficient of the squared term is zero. There are three roots for this equation and one can use complex numbers to see how they are related. Note that the sum of the three roots is zero.
Knowing the form of the solutions one can choose the components of the complex number x=r'+iq' and find the equation for which these numbers are the roots of. The equation below is the one that Cardan uses in Ars Magna.
The trick used to solve the simplified cubic equation involves substituting x=u-v and finding an expression relating u and v then using this to eliminate v and obtain a quadratic equation that is a function of u3. One can then use the quadratic formula to find u and the corresponding real root. The other roots can be found by multiplying u by the cube roots of 1 and solving for the root using the same method. The second solution for the quadratic equation involving u yields the same set of roots for the simplified cubic equation.
Supplemental (Mar 2): The above method for finding the roots of a cubic equation appears to work even if all the roots are real. For some choices of B the radical is imaginary and the associated roots end up as reals. For a double root (two roots the same) the radical is zero and for a triple root (all three roots the same) both B and C of the reduced equation are zero.
Knowing the form of the solutions one can choose the components of the complex number x=r'+iq' and find the equation for which these numbers are the roots of. The equation below is the one that Cardan uses in Ars Magna.
The trick used to solve the simplified cubic equation involves substituting x=u-v and finding an expression relating u and v then using this to eliminate v and obtain a quadratic equation that is a function of u3. One can then use the quadratic formula to find u and the corresponding real root. The other roots can be found by multiplying u by the cube roots of 1 and solving for the root using the same method. The second solution for the quadratic equation involving u yields the same set of roots for the simplified cubic equation.
Supplemental (Mar 2): The above method for finding the roots of a cubic equation appears to work even if all the roots are real. For some choices of B the radical is imaginary and the associated roots end up as reals. For a double root (two roots the same) the radical is zero and for a triple root (all three roots the same) both B and C of the reduced equation are zero.